merination notes

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BITSA

BITSAT P

T PAPER

APER

09.09.2012 (PT-02)

PHYSICAL CHEMISTRY

ORGANIC CHEMISTRY TEST SYLLABUS

Reaction Mechanism :

Reaction Mechanism : SN1 of RSN1 of R – –X, RX, R – –OO – –R & R & RR – –OH, E1 in Alcohols, SOH, E1 in Alcohols, S_N_N2 of R2 of R – –X, RX, R – –OO – –R & R & RR – –OH, Sni, Intramo-OH, Sni,

Intramo-lecular S

lecular S_N_N2, 2, Sayt Zeff product, Sayt Zeff product, Hoffmann product, E1 CB, Comparison Hoffmann product, E1 CB, Comparison of E2/E1CB of E2/E1CB Physical PropertPhysical Propertiesies of Alkyl halide, Alcohol & Ether.

of Alkyl halide, Alcohol & Ether. Physical SCQ (32)

Physical SCQ (32) 1.

1. A solution is a mixture of A solution is a mixture of 0.06 M KCl and 0.06 M KCl and 0.06 M KI. AgNO0.06 M KI. AgNO₃₃ solution is solution is being added drop by drop till AgClbeing added drop by drop till AgCl starts precipitating (K

starts precipitating (K_sp_sp AgCl = 1 x 10 AgCl = 1 x 10!!1010_{and K}_{and K} sp

sp AgI = 4 x 10 AgI = 4 x 10 !

!1616_{). The concentration of iodide ion at}_{). The concentration of iodide ion at this instant}_{this instant}

will be nearly equal to : will be nearly equal to : (A) 4.0 x 10

(A) 4.0 x 10!!55_M_M _((B_B)₎₄₄_x_x₁₁₀₀!!88_M_M _((C_C)₎_2..4₂₄_x_x₁₁₀₀!!88_M_M _((D_D**)_{) 2}_2..4_{4 x}_{x 1}₁₀₀!!77_M_M

Sol.

Sol. When When AgCAgCl Sl Statarts rts preprecipcipitaitatintingg [Ag[Ag++_{] =}_{] =}

06 06 .. 0 0 10 10!!1010

at that time conc of [S at that time conc of [S – –_{] =}_{] =}

10 10 16 16 10 10 10 10 4 4 ! ! ! ! " " = 2.4 = 2.4 × 10× 10 – –77 2.

2. Heat of neutralization of NHHeat of neutralization of NH₄₄OH and HCl isOH and HCl is

((AA) ) 1133..7 7 kkccaall//mmoollee ((BB**) ) < 1< 133..7 7 kkccaall//mmoollee ((CC) ) > > 1133..7 7 kkccaall//mmoollee ((DD) ) ZZeerroo Sol.

Sol. NHNH₄₄OH is a weak base. Heat of netralisation < 13.7 kcal.OH is a weak base. Heat of netralisation < 13.7 kcal. 3.

3. The solubility of AgCl will be minimum inThe solubility of AgCl will be minimum in (A) 0.001 M AgNO

(A) 0.001 M AgNO₃₃ (B) Pure water(B) Pure water

a a

(C*) 0.01 M CaCl(C*) 0.01 M CaCl₂₂ (D) 0.01 M NaCl(D) 0.01 M NaCl Sol.

Sol. 0.01 M CaCl0.01 M CaCl₂₂ gives maximum Cl gives maximum Cl – –_{ions. To keep K}_{ions. To keep K}

sp

sp of AgCl constant, decrease in of AgCl constant, decrease in [Ag[Ag +

+_{] will be maximum}_{] will be maximum}

4.

4. A weak base BOH A weak base BOH (0.1 mole) is titrated w(0.1 mole) is titrated with strong acid HCl (0.08 mole) thaith strong acid HCl (0.08 mole) than the number n the number of Hof H++_{ion is (K}_{ion is (K} b bforfor

BOH = 10 BOH = 10 – –44₎₎

(A*) 24.08

(A*) 24.08""₂₀₂₀ – –33 _{(B) 4}_{(B) 4}_"_"₁₀₁₀ – –1010 _{(C) 6.02}_{(C) 6.02}_"_"₁₀₁₀1313 _{(D) None}_{(D) None}

Sol.

Sol. BOBOHH + + HHCCll # # #$#$ BCl + HBCl + H₂₂OO

t

t = = 00 00..1 1 mmoollee 00..008 8 mmoollee

tt= e= eqq.. 00..002 2 momollee –– 0.08 mole0.08 mole

%

% as solution is buffer as solution is buffer so so [OH [OH – –_{] = 10}_{] = 10} – –44_"_" 08 08 .. 0 0 02 02 .. 0 0 [OH [OH – –_{] = 10}_{] = 10} – –44_"_" 4 4 1 1 mole mole % % [H [H++_{] =}_{] =} 4 4 14 14 10 10 4 4 10 10 ! ! ! ! _"_" [H [H++_{] = 4}_{] = 4}_"_"₁₀₁₀ – –1010 %

% No. of H No. of H++_{ion = 4}_{ion = 4}_"_"₁₀₁₀ – –1010_"_"_6.02_6.02_"_"₁₀₁₀2323

= 24.08

= 24.08""₁₀₁₀+13+13

5.

5. A certain buffer solution contains XA certain buffer solution contains X – –_{and HX}_{and HX with their concentrations related as}_{with their concentrations related as} ]] X X [[ ]] HX HX [[ ! ! = 0.2 = 0.2 If the value of K

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Page # 2 Page # 2 5.

If the value of K_b_b at 25 at 25°C for X°C for X – – is 10 is 10 – –99, pH of the buffer at 25, pH of the buffer at 25°C is : (log 2 = 0.3)_{°C is : (log 2 = 0.3)} ((AA**) ) 55..77 ((BB) ) 88..33 ((CC) ) 99..77 ((DD) ) 44..33 Sol.

Sol. KK_b_b for X for X – –_{= 10}₌₁₀ – –99 _&_&_K_K

a a for HX = 10 for HX = 10 – –55 _&_&_pK_pK a a = 5 = 5 pH = pK pH = pK_a_a + log + log₁₀₁₀ ]] HX HX [[ ]] X X [[ !! = 5 = 5 + 0.7 = + 0.7 = 5.75.7 6. 6. The pH of aThe pH of a 20 20 M M

solution of a weak base, if its K

solution of a weak base, if its K_b_b value at 25 value at 25°C is 2.5°C is 2.5""₁₀₁₀ – –33, will be : [Given log 11.18 = 1.05], will be : [Given log 11.18 = 1.05]

((AA) ) 1111..9955 ((BB**) ) 1122 ((CC) ) 1122..0055 ((DD) ) 11..9955 Sol. Sol. KK_b_b = = _!_!''_'_' 1 1 C C 22 & & ''_{= 0.2}_{= 0.2} &

&_[OH_[OH – –] = C] = C_'_'= 0.05= 0.05_"_" 0.2 = 0.01 0.2 = 0.01 _&_& pOH = 2 pOH = 2 _&_& pH = 14 pH = 14_–_–_{2 = 12}_{2 = 12} 7.

7. If equal volume of following solutions are mixed, precipitation of HgIf equal volume of following solutions are mixed, precipitation of Hg₂₂II₂₂(K(K_sp_sp = 2.5 = 2.5""1010 – –2626_{) will occur onl}_{) will occur only with :}_{y with :}

(A) 10 (A) 10 – –44_M_M 22)) 2 2 Hg Hg + 10 + 10 – –111111_M_M I I – – _{(B) 10}_{(B) 10} – –55_M_M 22)) 2 2 Hg Hg + 10 + 10 – –1010_M_M_I_I – – (C) 10 (C) 10 – –1111_M_M 22)) 2 2 Hg Hg + 10 + 10 – –77_M_M I I – – (D*) 10(D*) 10 – –66 M M HgHg₂₂22)) + 10 + 10 – –99 M MII – – Sol. Sol. 1010 – –66_M_{M Hg}_Hg 2 2 2+ 2+₊_{+ 10}₁₀ – –99_M_{M II} – – & &_{IP =}_{IP =}

2 2 10 10!!66 " " 2 2 9 9 2 2 10 10 ** ** + + , , --. . / / !! = = 8 8 10 10!!2424 = 1.25 = 1.25 ""₁₀₁₀ – –2525_{> Ksp}_{> Ksp} &

&_{precipitation of Hg}_{precipitation of Hg}

2

2II22 will occur. will occur.

8.

8. In which of the following solutions, the In which of the following solutions, the degree of dissociation of Hdegree of dissociation of H₂₂O is less than 1.8O is less than 1.8 ""₁₀₁₀ – –77_{% at 25}_{% at 25° C :}_{° C :}

(A) 10

(A) 10 – –66_M_M_H_HC_Cll _((B_B)₎₁₁₀₀ – –77_M_M_N_Na_aO_OH_H _((C_C)_{) 1}₁₀₀ – –88_M_{M H}_HC_Cll _((D_D**)_{) A}_Alll_{l o}_of_{f tth}_he_es_se_e

Sol.

Sol. The degree of dissociation of pure water at 25The degree of dissociation of pure water at 25°C = 1.8°C = 1.8""₁₀₁₀ – –77%%

&

&_{any H}_{any H}++_{or OH}_{or OH} – –_{ions from an}_{ions from an external source will suppress the dissociation of H}_{external source will suppress the dissociation of H}

2 2O.O.

9.

9. Three sparingly soluble salts MThree sparingly soluble salts M₂₂X, MX and MXX, MX and MX₃₃ have their solubility prod have their solubility product in the ratio of uct in the ratio of 4: 1 : 27. Their4: 1 : 27. Their solubilities will be in the order :

solubilities will be in the order : (A) MX (A) MX₃₃ > MX > M > MX > M₂₂XX ((BB**) ) MMXX₃₃> > MM₂₂X X > > MMXX ((CC) ) MMX X > > MMXX₃₃> > MM₂₂XX ((DD) ) MMX X > > MM₂₂X > MXX > MX₃₃ Sol. Sol. FFoorr MM₂₂X X ,, 44SS₁₁33₌₌₄_4x_x_;; _S_S 1 1 = x = x 1/3 1/3 F Foorr MMX X ,, 44SS₂₂22₌₌_x_x _;; _S_S 2 2 = x = x 1/2 1/2 F Foorr MMXX₃₃ , , 2277SS₃₃44₌_{= 2}₂₇_7x_{x ;;} _S_S 3 3 = x = x 1/4 1/4 & & _S_S 3 3 > S > S11 > S > S22 10.

10. Calculate the pH of a 0.1 M KCalculate the pH of a 0.1 M K₃₃POPO₄₄ solution. The third dissociation constant of phosphoric acid is 10 solution. The third dissociation constant of phosphoric acid is 10 – –1212_..

Given (0.41)

Given (0.41)1/21/2 _{= 0.64}_{= 0.64 ; log 3}_{; log 3 = 0.48}_{= 0.48}

( (AA) ) 1122..55 ((BB**) ) 1122..4444 ((CC) ) 1122..2255 ((DD) ) 1122 S Sooll.. KK_h_h = = 3 3 a a w w K K K K = = ₁₂₁₂ 14 14 10 10 10 10 ! ! ! ! = 10 = 10 – –22 _%_% KK h h = = ₍₍₁₁ _h_h₎₎ Ch Ch22 ! !

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Page # 2 Page # 2 5.

If the value of K_b_b at 25 at 25°C for X°C for X – – is 10 is 10 – –99, pH of the buffer at 25, pH of the buffer at 25°C is : (log 2 = 0.3)_{°C is : (log 2 = 0.3)} ((AA**) ) 55..77 ((BB) ) 88..33 ((CC) ) 99..77 ((DD) ) 44..33 Sol.

Sol. KK_b_b for X for X – –_{= 10}₌₁₀ – –99 _&_&_K_K

a a for HX = 10 for HX = 10 – –55 _&_&_pK_pK a a = 5 = 5 pH = pK pH = pK_a_a + log + log₁₀₁₀ ]] HX HX [[ ]] X X [[ !! = 5 = 5 + 0.7 = + 0.7 = 5.75.7 6. 6. The pH of aThe pH of a 20 20 M M

solution of a weak base, if its K

solution of a weak base, if its K_b_b value at 25 value at 25°C is 2.5°C is 2.5""₁₀₁₀ – –33, will be : [Given log 11.18 = 1.05], will be : [Given log 11.18 = 1.05]

((AA) ) 1111..9955 ((BB**) ) 1122 ((CC) ) 1122..0055 ((DD) ) 11..9955 Sol. Sol. KK_b_b = = _!_!''_'_' 1 1 C C 22 & & ''_{= 0.2}_{= 0.2} &

&_[OH_[OH – –] = C] = C_'_'= 0.05= 0.05_"_" 0.2 = 0.01 0.2 = 0.01 _&_& pOH = 2 pOH = 2 _&_& pH = 14 pH = 14_–_–_{2 = 12}_{2 = 12} 7.

7. If equal volume of following solutions are mixed, precipitation of HgIf equal volume of following solutions are mixed, precipitation of Hg₂₂II₂₂(K(K_sp_sp = 2.5 = 2.5""1010 – –2626_{) will occur onl}_{) will occur only with :}_{y with :}

(A) 10 (A) 10 – –44_M_M 22)) 2 2 Hg Hg + 10 + 10 – –111111_M_M I I – – _{(B) 10}_{(B) 10} – –55_M_M 22)) 2 2 Hg Hg + 10 + 10 – –1010_M_M_I_I – – (C) 10 (C) 10 – –1111_M_M 22)) 2 2 Hg Hg + 10 + 10 – –77_M_M I I – – (D*) 10(D*) 10 – –66 M M HgHg₂₂22)) + 10 + 10 – –99 M MII – – Sol. Sol. 1010 – –66_M_{M Hg}_Hg 2 2 2+ 2+₊_{+ 10}₁₀ – –99_M_{M II} – – & &_{IP =}_{IP =}

2 2 10 10!!66 " " 2 2 9 9 2 2 10 10 ** ** + + , , --. . / / !! = = 8 8 10 10!!2424 = 1.25 = 1.25 ""₁₀₁₀ – –2525_{> Ksp}_{> Ksp} &

&_{precipitation of Hg}_{precipitation of Hg}

2

2II22 will occur. will occur.

8.

8. In which of the following solutions, the In which of the following solutions, the degree of dissociation of Hdegree of dissociation of H₂₂O is less than 1.8O is less than 1.8 ""₁₀₁₀ – –77_{% at 25}_{% at 25° C :}_{° C :}

(A) 10

(A) 10 – –66_M_M_H_HC_Cll _((B_B)₎₁₁₀₀ – –77_M_M_N_Na_aO_OH_H _((C_C)_{) 1}₁₀₀ – –88_M_{M H}_HC_Cll _((D_D**)_{) A}_Alll_{l o}_of_{f tth}_he_es_se_e

Sol.

Sol. The degree of dissociation of pure water at 25The degree of dissociation of pure water at 25°C = 1.8°C = 1.8""₁₀₁₀ – –77%%

&

&_{any H}_{any H}++_{or OH}_{or OH} – –_{ions from an}_{ions from an external source will suppress the dissociation of H}_{external source will suppress the dissociation of H}

2 2O.O.

9.

9. Three sparingly soluble salts MThree sparingly soluble salts M₂₂X, MX and MXX, MX and MX₃₃ have their solubility prod have their solubility product in the ratio of uct in the ratio of 4: 1 : 27. Their4: 1 : 27. Their solubilities will be in the order :

solubilities will be in the order : (A) MX (A) MX₃₃ > MX > M > MX > M₂₂XX ((BB**) ) MMXX₃₃> > MM₂₂X X > > MMXX ((CC) ) MMX X > > MMXX₃₃> > MM₂₂XX ((DD) ) MMX X > > MM₂₂X > MXX > MX₃₃ Sol. Sol. FFoorr MM₂₂X X ,, 44SS₁₁33₌₌₄_4x_x_;; _S_S 1 1 = x = x 1/3 1/3 F Foorr MMX X ,, 44SS₂₂22₌₌_x_x _;; _S_S 2 2 = x = x 1/2 1/2 F Foorr MMXX₃₃ , , 2277SS₃₃44₌_{= 2}₂₇_7x_{x ;;} _S_S 3 3 = x = x 1/4 1/4 & & _S_S 3 3 > S > S11 > S > S22 10.

10. Calculate the pH of a 0.1 M KCalculate the pH of a 0.1 M K₃₃POPO₄₄ solution. The third dissociation constant of phosphoric acid is 10 solution. The third dissociation constant of phosphoric acid is 10 – –1212_..

Given (0.41)

Given (0.41)1/21/2 _{= 0.64}_{= 0.64 ; log 3}_{; log 3 = 0.48}_{= 0.48}

( (AA) ) 1122..55 ((BB**) ) 1122..4444 ((CC) ) 1122..2255 ((DD) ) 1122 S Sooll.. KK_h_h = = 3 3 a a w w K K K K = = ₁₂₁₂ 14 14 10 10 10 10 ! ! ! ! = 10 = 10 – –22 _%_% KK h h = = ₍₍₁₁ _h_h₎₎ Ch Ch22 ! !

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Page # 3 Page # 3 S Sooll.. KK_h_h = = 3 3 a a w w K K K K = = ₁₂₁₂ 14 14 10 10 10 10 ! ! ! ! = 10 = 10 – –22 _%_% _K_K h h = = ₍₍₁₁ _h_h₎₎ Ch Ch22 ! ! as 1 as 1 – – h = 1, h = h = 1, h = C C K K_h_h = = 1 1 .. 0 0 10 10!!22 = 0.316 = 0.316 as as h h > > 0.10.1 %% 1 1 – – h h 00 1 1 & & 10 10 – –22₌₌ )) h h 1 1 (( h h 10 10 11 22 ! ! " " ! ! or 0.1 (1 or 0.1 (1 – – h) = h h) = h22 or, 0.1 or, 0.1 – – 0.1 h = h 0.1 h = h22 or, or, h h + 0.1 + 0.1 hh – – 0.1 = 0 0.1 = 0 or, h = or, h = 2 2 1 1 .. 0 0 4 4 )) 1 1 .. 0 0 (( 1 1 .. 0 0 )) 22 )) "" ! ! = 0.27 = 0.27 as, PO as, PO₄₄33 – –₊_{+ H}_H 2 2O O HPOHPO44 2 2 – –₊_{+ OH}_OH – – c(1 c(1 – – h) h) cchh cchh &

& [OH [OH – –_{] = ch}_{] = ch} = 0.1

= 0.1 ""_0.27_0.27

= 27

= 27 ""₁₀₁₀ – –33

pOH = 3

pOH = 3 – – log 27 = 3 log3 log 27 = 3 log333 = 3 = 3 – – 3 log 3 3 log 3

= 3

= 3 – – 3 3"" 0.48 = 1.56 0.48 = 1.56

pH = 14

pH = 14 – – 1.56 = 12.44 1.56 = 12.44

1

11.1. The pKThe pK_a_a of HCN is 9.3. Tof HCN is 9.3. The pH of a solution prepared by mixture 2.5 mole of he pH of a solution prepared by mixture 2.5 mole of KCN and 2.5 mole of HCN KCN and 2.5 mole of HCN inin water and making up the total volume to 500 ml

water and making up the total volume to 500 ml isis

((AA**) ) 99..33 ((BB) ) 77..33 ((CC) ) 1100..33 ((DD) ) 88..33 Sol.

Sol. pH = pKpH = pK_a_a + log + log _[[[[_Acid_AcidSaltSalt]]_]] = 9.3 = 9.3 12.

12. Calculate the molar solubility of AgCl in 2.5 Calculate the molar solubility of AgCl in 2.5 M NHM NH₃₃ solution. [Given : KspAgCl = 10 solution. [Given : KspAgCl = 10 – –1010_{, K}_{, K}

ff[Ag(NH[Ag(NH33))22]] +

+₌_{= 10}₁₀66_]]

((AA**) ) 00..00225 5 mmooll//LL ((BB) ) 00..2 2 L L mmooll – –11 _{(C) 0.4 L mol}_{(C) 0.4 L mol} – –11 _{(D) None of these}_{(D) None of these}

Sol.

Sol. AgCl(s) AgCl(s) AgAg++_{(aq) + Cl}_{(aq) + Cl} – –_(aq)_(aq)

Ag Ag++_{+ 2NH}_{+ 2NH} 3 3 [Ag(NH[Ag(NH33))22]] + + [Ag [Ag++_{] =}_{] =} s s K K_sp_sp .... (1) .... (1) K K_ff = = _[[_Ag_Ag _](_](₂₂_..₅₅ ₂₂_s_s₎₎22 s s ! ! ) ) or, Kor, K ff == 22 sp sp 2 2 )) s s 2 2 5 5 .. 2 2 (( K K s s ! ! or, Kor, Kff"" K Kspsp = = 22 2 2 )) s s 2 2 5 5 .. 2 2 (( s s ! ! or, or, 1010 – –22 = = 22..55 22ss s s ! ! or, or, s s 2 2 5 5 .. 2 2 s s !

! = 10 = 10 – –1212 or, or, s s = = 0.0250.025 – – 0.02 s 0.02 s or, or, 1.02 1.02 s s = = 0.025 0.025 or, or, s s == 11..0202 025 025 .. 0 0 1 1 0.025 0.025 mol/Lmol/L 13.

13. What is the concentration of acetic acid which can be added to 0.5 M formic acid so that the % dissociationWhat is the concentration of acetic acid which can be added to 0.5 M formic acid so that the % dissociation of neither acid is changed by the addition. K

of neither acid is changed by the addition. K_a_a for acetic acid is 1.85 for acetic acid is 1.85 × 10× 10-5-5_{, K}_{, K} a

a for formic acid = for formic acid = 2.42.4 × 10× 10 -4 -4_..

((AA) ) AAnny y ccoonncceennttrraattiioonn ((BB) ) TThheerre e ccaan n nnoot t bbe e aanny y ccoonncceennttrraattiioonn ((CC**) ) 66..666 6 MM ((DD) ) 33..333 3 MM Sol. Sol. CC₁₁'' 1 1 = = CC22''22 1 1 a a CC K K ₁₁ = = KKaa₂₂CC22 1.8 1.8 × 10× 10 – –55_{× C}_{× C} 1 1 = = 2.42.4 × 10× 10 – –44_{× 0.5}_{× 0.5} C C₁₁ = 6.66 M = 6.66 M 14. CH₃NH₂ (K_b= 5 × 10 –4_{) 0.1 mole of CH}

3NH2(Kb= 5 × 10 –4) is mixed with 0.08 mole of HCl and diluted to one

litre. What will be the H+_{concentration in the solution? is mixed with 0.08 mole of HCl and diluted to one litre.}

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Page # 4 14. CH₃NH₂ (K_b= 5 × 10 –4_{) 0.1 mole of CH}

3NH2(Kb= 5 × 10 –4) is mixed with 0.08 mole of HCl and diluted to one

litre. What will be the H+_{concentration in the solution? is mixed with 0.08 mole of HCl and diluted to one litre.}

What will be the H+_{concentration in the solution?}

(A) 8 × 10 –2_M _(B*)_{8 × 10} –11_M _(C)_{1.6 × 10} –11_M _(D)_{8 × 10} –5_M Sol. CH₃NH₂ + HCl # #$ CH₃_N_H₃ 2 l C 3 .1 .08 .08 .02 0 .08

For buffer sol. |OH –_{| = K}

b x _[_Salt_] ] Base [ = 5 x 10 –4_x 08 . 02 . |OH –_{| =} 4 5 x 10 –4 so |H+_{| =} 4 14 10 x 5 4 10 ! ! = 5 4 x 10 –10_{= 8 x 10} –111_{M Ans.}

15. 10 –2 mole of NaOH was added to 10 litre of water. The pH will change by

(A*)4 (B)3 (C)11 (D)7

Sol. Initially pH = 7

finally [NaOH] = 10 –3 _{so pOH = 3}

pH = 11 So, 4_{(pH) = 4}

16. The sum of negative logarithm of hydrogen ion and hydroxide ion concentration at 37 ºC : [K_w = 2.5 × 10 –4] (A) 14 (B*) Less than 14 (C) greater than 14 (D) Data insufficient.

Sol. pH + pOH = pK_W = 15 – log (5)2

= 15 – 2 × .699

= 13.6

17. In the reaction : [Ag(CN)₂] –+ Zn_#_#$ the complex formed will be :

(A*) Tetrahedral (B) square planar (C) octahedral (D) triangal bipyramidal Sol. 2 [Ag(CN)₂] –_{+ Zn}_#_#$_[Zn(CN)

4]

2 –_{+ 2 Ag} Tetrahedral

18. All the following complexes show a decreases in their weights when placed in a magnetic balance. Then which of the these has square planar geometry :

(A) Ni(CO)₄ (B*) K[AgF₄] (C)Na₂[Zn(CN)₄] (D) None of these Sol. K [AgF₄] is square planar because Ag(555_{) is 4d}8_{and complex is diamagnetic.}

19. It is an experiment fact that : DMG + Ni(55_{)salt + NH}

4OH # #$Red ppt.

Which of the following is wrong about this red ppt :

(A) It is a non –ionic complex (B) It involves intra molecular H –bonding

(C*) Ni(55_{) is sp}3_hybridised _{(D) It is a diamagnetic complex}

Sol. The complex is

20. Sodium nitroprusside is a diamagnetic substance and a important laboratory reagent for the testing of sulphide ions. The metal involved in the complexation in this is present in which of the following hybridisation state :

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Page # 5 20. Sodium nitroprusside is a diamagnetic substance and a important laboratory reagent for the testing of

sulphide ions. The metal involved in the complexation in this is present in which of the following hybridisation state :

(A) sp3 _{(B) dsp}2 _{(C*) d}2_sp3 _{(D) sp}3_d2

Sol. Sodium nitroprusside is Na₂ [Fe(CN)₅ (NO)

)

] ; a diamagnetic complex. 21. All the following complex ions are found to be paramagnetic :

P : [FeF₆]3 – _; _Q_:_[CoF 6] 3 – R : [V(H₂O)₆]3+ _; _S_:_[Ti(H 2O)6] 3+

The correct order of their paramagnetic moment (spin only) is :

(A*) P > Q > R > S (B) P < Q < R < S (C) P = Q = R = S (D) P > R > Q > S Sol. On the basis of number of electrons the correct order is P > Q > R > S.

22. When the complex K₆ [(CN)₅Co –O –O –Co(CN)₅] is oxidised by bromine into

K₅[(CN)₅ Co –O –O –Co(CN)₅]. Then which of the following statements will be true about this change:

(A) Co(55_{) is oxidised in Co(}555₎ _(B)_The_O_–_{O bond length will increase}

(C*) The O –O bond length will decrease (D) A‘ ’ &‘B’ both are correct

Sol. In the first complex ligand is O₂2 –_{which is oxidised into O}

2 1 –_. hence

O – O bond length decreases.

23. The octahedral complex [Rh(NO₂) (SCN) (en)₂]+_{can exist in a total number of isomeric forms including}

stereoisomers : (A)2 (B)4 (C)8 (D*)12 Sol. (1) NO₂ /SCN (5)NO₂ /SCN (9) (2) ONO / SCN (6) ONO / SCN (10) (3) NO₂ /NCS (7)NO₂ /NCS (11) (4) ONO NCS (8) ONO / NCS (12)

24. For the reaction Ni2+_{+ 4NH}

3 [Ni(NH3)4] 2+

at equilibrium, if the solution contains 1.6 × 10 –4% of nickel in the free state, And the concentration of NH

3at

equilibrium is 0.5 M. Then the instability constant of the complex will be approximately equal to : (A) 1.0 × 10 –5 _{(B) 1.5 × 10} –16 _{(C*) 1.0 × 10} –7 _{(D) 1.5 × 10} –17 Sol. Ni2+_{+ 4 NH} 3 [Ni(NH3)4] 2+ & _{k =} ₄ 3 2 2 4 3 ] NH [ ] Ni [ ] ) NH ( Ni [ ) ) But ) ) ) ) 2 4 3 2 2 ] ) NH ( Ni [ ] Ni [ ] Ni [ = 1.6 × 10 –6 or ) ) 2 4 3 2 ] ) NH ( Ni [ Ni 1_{1.6 × 10} –6 & _{k =} ₄ 6 ) 5 . 0 ( 6 . 1 10 " = 107

Hence instability constant = 10 –7

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Page # 6 25. In which of the following complex ion, the metal ion will have t6₂_g,e0_g configuration according to CFT::

(A) [FeF₆]3 – _{(B) [Fe(CN)}

6]

3 – _{(C*) [Fe(CN)}

6]

4 – _{(D) None of these} Sol. In [Fe(CN)₆]4 –_{; Fe(}₅₅_{) is t}

2g

6_{, eg}0_{due to strong ligands.}

26. Spin only magnetic moment of a complex having CFSE = – 0.64₀ and surrounded by weak field ligands can

be

(A) 1.73 BM (B) 4.9 BM (C*) both (A) & (B) (D) None of these Sol. The options can give CFSE = – 0.64₀ with weak field ligands% d4 and d9.

27. Which of the following statements is not correct? (a) [Ni(H₂O)₆]2+_{and [Ni(NH}

3)6]

2+_{have same value of CFSE}

(b) [Ni(H₂O)₆]2+_{and [Ni(NH} 3)6]

2+_{have same value of magnetic moment}

(A*) Only a (B) Only b (C) Both a and b (D) None of these Sol. Ammonia is a stronger field ligand than water.

28. The correct IUPAC name of the complex:

C = N OH OH H3C C = N H3C • • • • CoCl2 is :

(A*) Dichlorodimethylglyoximatecobalt (II) (B) Bis(dimethyglyoxime)dichlorocobalt (II) (C) Dimethylglyoximecobalt(II) chloride (D) Dichlorodimethylglyoxime-N, N-cobalt (II) 29. Which of the following pair of complexes have the same EAN of the central metal atoms/ions?

(A) [Cu(NH₃)₄]SO₄ and K₃[Fe(CN)₆] (B)K₄[Fe(CN)₆] and [Co(NH₃)₆]Cl₃ (C) K₃[Cr(C₂O₄)₃] and [Ni(CO)₄] (D*) all of the above

Sol. (A) [Cu(NH₃)₄]2+ _{= 29}

– 2 + 8 = 35 [Fe(CN)₆]3 – = 26_–_{3 + 12 = 35} (B) [Fe(CN)₆]4 – = 26_–_{2 + 12 = 36} [Co(NH3)6]3+ = 27 – 3 + 12 = 36 (C) [Cr(C₂O₄)₃]3 –_{= 27}_–_{3 + 12 = 36} [Ni(CO)₄] = 28 + 8 = 36

30. In the reaction [CoCl₂(NH₃)₄]+_{+ Cl} – _#_#$_[CoCl

3(NH3)3] + NH3 only one isomer of product is obtained .

Hence the initial complex must be

(A) cis isomer (B*) trans isomer (C) both (D) mixture of both Sol. Moderate

# # $ # Cl!

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Page # 7

# # $

# Cl! two isomers product

replacable positions 31. 3 AgNO . Aq_# # $ # ) Major ( 'X' product : (A) (B) (C*) (D)

Sol. Aqueous AgNO₃ catalyse S_N1 reaction. 32. Consider the following reaction.

ether SOCl_#_#2

# $

#

In the above reaction which phenomenon will take place :

(A) Inversion (B*) Retention (C) Racemisation (D) Isomerisation

Sol. It is S_Ni reaction so retention takes place

33. Which one of the following has maximum nucleophilicity ?

(A*) CH₃S3 (B) (C) Et₃N (D)

Sol. Nucleophilicity 6size (in a group).

34. # # NaCN # # $

In the given reaction rate is fastest, when (X) is :

(A) –OH (B) –NH₂ (C) O || CH O S || O 3 ! ! _(D*) O || CH S O || O 3 ! ! !

Sol. Leaving group ability6_{Stability of anion.}

35. In the following reaction the most probable product will be :

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Page # 8

(A) (B*) (C) (D)

Sol. 7

36. When the concentration of alkyl halide is tripled and the concentration of ion is reduced to half, the rate of S_N2 reaction increases by:

(A) 3 times (B) 2 times (C*) 1.5 times (D) 6 times Sol. Rate of S_N2' _[R_–_{X ] [Nu} –_]

1 2 r r = ] OH [ ] RX [ OH ] RX 3 [ – – 2 1 r₂= 1.5 r₁

37. In which of the following reaction the product obtained is t-butyl methyl ether ?

(A) CH₃OH + HO—CH₂—CH₃ # # conc # # .H2 # SO # # 4$ (B)

(C*) (D)

Sol. t-butyl methyl ehter is a mixed ether and for the preparation of mixed ethers in high yield the essential condition is the use of primary alkyl halide.

Thus,

#

# $

!NaBr

This reaction is williamson's synthesis.

38. CH₃CH₂CH₂OH # # PCl # # 5$ _A # # alc # KOH # # $_B

B is identified as :

(A) propanal (B) propane (C) propyne (D*) propene

Sol. ROH HCl / ZnCl or I) Br, , Cl X ( PX or SOCl or PCl 2 3 2 5 : # # # # # # # $ # _RCl CH₃CH₂CH₂OH HCl PCl₅ ! # # $ # 2 Alkene 3 CH CH CH : B is an alkene (propene)

39. The only alcohol that cannot be prepared by the indirect hydration of alkene is :

(A) ethyl alcohol (B) propyl alcohol (C) isobutyl alcohol (D*) methyl alcohol

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Page # 9 39. The only alcohol that cannot be prepared by the indirect hydration of alkene is :

(A) ethyl alcohol (B) propyl alcohol (C) isobutyl alcohol (D*) methyl alcohol

Sol. Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbons so alcohol of at least two carbon atoms can be formed.

H₂C = CH₂ ) SO H ( H 4 2 # # $ # ) ) C H | C H 2 3 # # # $ # ! 4 HSO H OSO CH| C H 3 2 3 # # $ # HOH Ethanol2 4 2SO CHCH OH H ) H₂C = CH – CH₃ rule) ff's Markowniko ( SO H HOH 4 2 # # $ # product Major 3 3 OH| CH CH C H ! ! product Minor 3 2 2 CH CH CH HO! ! )

40. Lucas reagent reacts fastest with :

(A) butanol –1 (B) butanol –2 (C*) 2 –methyl –propanol –2 (D) 2 –methyl –propanol –1

Sol. The order of reactivity with alcohols with lucas reagent is -3º > 2º > 1º

&_{Lucas reagent reacts fastest with 3º alcohol.}

(a)

(b)

(c)

(d)

&_{choice (C) is the answer as it is 3 º alcohol and rate of reaction is f astest for 3º alcohol.}

BITSAT- PT - 2 - XII - (09-09-12)

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Page # 1

BITSAT- PT - 2 - XII - (09-09-12)

Syllabus : Sequence&Series, P&C, Binomial Theorem, M athematical Induction, Determinant, Straight lines 1. For every natural number n, n(n + 3) is always

(A*) even (B) odd (C) multiple of 4 (D) multiple of 5 Sol. Let P(n) = n(n + 3), then

P(1) = 1(4) = 4 which is even and multiple of 4. P(2) = 2(5) = 10 which is even and multiple of 5. P(3) = 3(6) = 18 which is even.

Hence it is clear that P(n) is even ! n" N 2. The greatest positive integer which divides 32n

– 2n – 1 ! n " N is (A)1 (B*) 2 (C)4 (D)8 Sol. Let P(n) = 32n – 2n – 1, then P(1) = 32 _–₂_–_{1 = 6} P(2) = 34 – 4 – 1 = 76 P(3) = 36 – 6 – 1 = 722

Obviously 2 is the greatest positive integer which divides P(n) ! n" N. 3. If the 9th terms of an A.P. be zero, then the ratio of its 29th and 19th term is

(A) 1 : 2 (B*) 2 : 1 (C) 1 : 3 (D) 3 : 1 Sol. Given that 9th term = a + (9 – 1) d = 0 $ a + 8d = 0

Now ratio of 29th and 19th terms = d 18 a d 28 a % % = d 10 ) d 8 a ( d 20 ) d 8 a (

%

= d 10 d 20 = 1 2

4. The solution of the equation (x + 1) + (x + 4) + (x + 7) + ....+ (x + 28) = 155 is

(A*) 1 (B) 2 (C) 3 (D) 4

Sol. We have (x + 1) + (x + 4) + (x + 7) + ....+ (x + 28) = 155 Let n be the number of terms in the A.P. on L.H.S. Then x + 28 = (x + 1) + (n – 1)3 $ n = 10 & (x + 1) + (x + 4) +...+ (x + 28) = 155 $ 2 10 [(x + 1) + (x + 28)] = 155 $ x = 1

5. If the arithmetic and geometric means of a and b be A and G respectively, then the value of A – G will be

(A) a b – a (B) 2 b a % (C*) 2 2 b – a ' ' ( ) * * + , (D) b a ab 2 %

Sol. Arithmetic mean of a and b = A = 2

b a %

and geometric mean G = _ab

Then A – G = 2 b a % – _{ab =} 2 ab 2 – b a% = 2 ) b )( a ( 2 – ) b ( ) a ( 2 % 2 = 2 2 b – a ' ' ( ) * * + ,

6. In series 1,2,2,2,2,3,3,3,3,3,3,3,3,3, 4,...the 400th term is

(A) 9 (B) 10 (C*) 11 (D)12

Sol. Number 11 starts at

-

₍₁2

_%

₂2

_%

_...

_%

₁₀2₎

_%

₁

.

th_position

i.e. 386th_position.

7. The sum of the series ...

3 3 1 3 2 2 1 2 1 1 1 1 4 2 4 2 4 2

_%

%

_%

%

_%

%

to n terms is (A) 1 n n ) 1 n ( n 2 2 % % % (B*) ) 1 n n ( 2 ) 1 n ( n 2 _% _% % (C) ) 1 n n ( 2 ) 1 n ( n 2 2 % % % (D) 2 n n2

%

Sol. Let T_n be the nth_{term of the series}

... 3 3 1 3 2 2 1 2 1 1 1 1 4 2 4 2 4 2

_%

%

_%

%

_%

%

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Page # 2 Sol. Let T_n be the nth_{term of the series}

... 3 3 1 3 2 2 1 2 1 1 1 1 4 2 4 2 4 2

_%

%

_%

%

_%

%

Then T_n = ₂ ₄ n n 1 n % % = (1 n2)2 –n2 n % = ₍_n _n ₁₎₍_n _–_n ₁₎ n 2 2 _% _% _% = '₍ ) * + , % % % _n _n ₁ 1 – 1 n – n 1 2 1 2 2 = ' ( ) * + , % % % 1 n(n 1) 1 – n ) 1 – n ( 1 1 2 1 Now 2 1 T n 1 r r

/

0 0 _' ( ) * + , %₁_.₂ 1 1 – 1 1 + 2 1 ' ( ) * + , % % ₁ ₂_.₃ 1 – 2 . 1 1 1 + 2 1 ' ( ) * + , % % ₁ ₃_.₄ 1 – 3 . 2 1 1 +...+ ' ( ) * + , % % % ₁ _n₍_n ₁₎ 1 – n ) 1 – n ( 1 1 2 1 = ' ( ) * + , % %n(n 1) 1 1 – 1 2 1 = ) 1 n n ( 2 ) 1 n ( n 2 _% _% %

8. The number of common terms to the sequence 17, 21, 25, ...417 and 16, 21, 26,...466 is

(A)21 (B) 19 (C*)20 (D) 22

Sol. Common terms are 21,41,61,...(d = LCM of 4,5 = 20) t_n1 417

$ 21 + (n – 1)201 417

$ _n 1_20.8

$ _{max value of n = 20}

9. If a, b, c are in A.P. as well as in G.P., then

(A) a = b 2 c (B)a 2 b = c (C) a 2 b2 c (D*) a = b = c Sol. As given b = 2 c a % ...(i) and b2_{= ac} $ _{(a + c)}2_{= 4ac} $ _(a_–_c)2_{= 0} $ _{a = c}

putting a = c in (i) we get b = c

& a = b = c

10. If G be the geometric mean of x and y, then 2 2 2 _y2

– G 1 x – G 1 _% = (A) G2 _(B*) 2 G 1 (C) ₂ G 2 (D) 3G2 Sol. As given G = xy & ₂ ₂ ₂ ₂ y – G 1 x – G 1 _% = 2 x – xy 1 + 2 y – xy 1 = _y – x 1 33 4 5 66 7 8 % y 1 x 1 – = xy 1 = ₂ G 1

11. If a₁, a₂, a₃,... are in A.P. such that a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 225, then a₁ + a₂ + a₃ + ... + a₂₃ + a₂₄ is equal to

(A) 909 (B) 75 (C) 750 (D*) 900

Sol. a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 225

$ 3 (a₁ + a₂₄) = 225 (sum of terms equidistant from beginning and end are equal) a₁ + a₂₄ = 75

Now a₁ + a₂ + ... + a₂₃ + a₂₄

= 2 24

[a₁ + a₂₄] = 12 × 75 = 900

12. There are n distinct points on the circumference of a circle. The number of pentagons that can be formed with these points as vertices is equal to the number of possible triangles. Then the value of n

is-(A)7 (B*)8 (C)15 (D)30

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Page # 3 12. There are n distinct points on the circumference of a circle. The number of pentagons that can be formed

with these points as vertices is equal to the number of possible triangles. Then the value of n

is-(A)7 (B*)8 (C)15 (D)30 Sol. nC =₅ nC₃ 5 n_{C =} 3 – n n_C 5 = n – 3 n = 8

13. A question paper consists of two parts A and B. Part A has 4 questions in which each question has an alternative and part B has 3 questions without any alternative. The number of ways to attempt paper when at least one question must be attempted for each part is (are)

(A) 561 (B*) 560 (C) 648 (D) 127

Sol. Required ways = (34_–₁₎₍₂3_–_{1) = 560}

14. Number of ways such that 6 boys and 3 girls can be seated such that there is exactly one boy in between any two girls

(A) 50400 (B*) 21600 (C) 10800 (D) 36000

Sol. Number of ways = 6! × 5 × 3!

15. All letters of the word 'RACHIT' are permuted in all possible ways and the words so formed (with or without meaning) are written as in dictionary, then the 484th_word

is-(A) RACHIT (B*) RACITH (C) RACTHI (D) RACIHT

Sol. ACHIRT $ 5!

CAHIRT $ 5!

HACIRT $ 5!

IACHRT $ 5!

RACHIT is 481th_word _RACHIT

RACHTI is 482th_word

RACIHI is 483th_word

RACITH is 484th_word

16. The number of ways in which 6 different red roses and 3 different white roses can form a garland so that all the white roses come together is

(A*) 2160 (B) 2165 (C) 2155 (D) 4320 Sol. ways = 2 ! 3 ! ) 1 – 7 ( 9 = 2160

17. There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through at least 3 of these points is

(A) 116 (B*) 117 (C) 120 (D) 115

Sol. Total number of solutions = 10_C 3 –

4_C

3 + 1 = 117.

18. How many different arrangements can be made out of the letters in the expansion A2_B3_C4_{, when written}

in full length ? (A*) ₂_!₃9_!!₄_! (B) 4 . 3 . 2 ! 9 (C) 2 ! 3 ! – 4 (D) ₂_! ₃_! ₄_! ! 9 % % Sol. Here A, B, C are repeated twice, thrice and four times respectively

& No. of arrangements = (2₂_!₃3_!₄_!4)!

% %

= ₂_!₃9_!!₄_!

19. Number of positive integral solutions of x₁. x₂ x₃ = 210

is-(A)25 (B) 26 (C)27 (D*)81

Sol. We have

x₁ x₂ x₃ = 210 = 2.3.5.7

& Total no. of solutions of the equation x₁x₂x₃ = 210 is 3 × 3 × 3 × 3 = 81

20. Total number of ways in which 15 identical blankets can be distributed among 4 persons so that each of them gets at least two blankets, equal to

(A*) 10_C 3 (B) 9_C 3 (C) 11_C 3 (D) none of these

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Page # 4 20. Total number of ways in which 15 identical blankets can be distributed among 4 persons so that each of

them gets at least two blankets, equal to (A*) 10_C 3 (B) 9_C 3 (C) 11_C 3 (D) none of these

Sol. Let 4 persons recieve B₁, B₂, B₃, B₄ number of blankets

& B₁ + B₂+ B₃ + B₄ = 15 {B₁,B₂, B₃ ,B₄: 2} Thus number of ways = 10_C

3

21. Let S (k) = 1 + 3 + 5 +...+ (2k – 1) = 3 + k2. Then which of the following is true ?

(A) S(1) is correct (B*) S(k) $ S (k + 1) (C) S(k) S(k + 1)

(D) Principle of mathematical induction can be used to prove the formula Sol. S(k) = 1 + 3 + 5 + ... + (2k – 1) = 3 + k2

put k = 1 in both sides, we get

& LHS = 1 and RHS = 3 + 1 = 4

$ _LHS2_RHS

Put (k + 1) in both sides in the place of k LHS = 1 + 3 + 5 + .... + (2k – 1) + (2k + 1) RHS = 3 + (k + 1)2_{= 3 + k}2_{+ 2k + 1} Let LHS = RHS 1 + 3 + 5 + ... + (2k – 1) + (2k + 1) = 3 + k2_{+ 2k + 1} $ _{1 + 3 + 5 + ... + (2k} – 1) = 3 + k2

If S(k) is true, then S(k + 1) is also true.

Hence, S(k) $ S(k + 1).

22. If the coefficients of second, third and fourth terms in the expansion of (1 + x)2n_{are in A.P., then which of the}

following is TRUE. (A) n2 – 9n + 7 = 0 (B) 3n2 – 9n + 7 = 0 (C) 3n2 + 9n + 7 = 0 (D*) 2n2 – 9n + 7 = 0 Sol. 2n_C 1 , 2n_C 2 , 2n_C 3 are in A.P. $ 2n2_–_{9n + 7 = 0}

23. The term that is independent of x in the expansion of

9 2 x 3 1 x 2 3 ₃ 4 5 6 7 8 ; _is (A) 4 5 6 9 3 1 2 3 C 3 4 5 6 7 8 ; 3 4 5 6 7 8 (B*) 3 3 9 6 1 C 3 4 5 6 7 8 (C) 5 4 4 9 3 1 2 3 C 3 4 5 6 7 8 ; 3 4 5 6 7 8 (D) 6 6 6 9 3 13 2 3 C 3 4 5 6 7 8 ; 3 4 5 6 7 8 Sol. 9 2 x 3 1 x 2 3 ₃ 4 5 6 7 8 ; ₌

/

0 ;

3

4

5

6

7

8

3

4

5

6

7

8

9 0 r r r 9 2 r 9 x 3 1 – x 2 3 C

For the term that is independent of x 18 – 2r – r = 0$ r = 6 Required term = 9C₆ 3 2 3₃ 4 5 6 7 8 6 3 1₃ 4 5 6 7 8 ; ₌ 6 9_C 3 6 1₃ 4 5 6 7 8 24. In the expansion of (21/5₊

3 )20, the sum of all rational terms is equal to

(A) 21 (B) 84 (C) 97 (D*) none of these

Sol. T_r+1 = 20C (2_r 1/5₎20 –r ( 3 )r = 20_{C .}_r 4 ₅r

2 ; . 2 r

3

As 2 and 3 are relatively primes. T_r+1 is rational, if 5 r and 2 r are integer’’ss & r is multiple of 10 & 01 r 1 20 r = 0 ,10, 20

Thus sum of rational terms = T₁ + T₁₁ + T₂₁ =20_C 0 2 4₊20_C 10 2 2_{. 3}5 ₊20_C 20 . 3 10

This is more than 21, 84, 97

25. The value of

_<=

<

>

?

<@

<

A

B

28 32003

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Page # 5 25. The value of

<=

<

>

?

<@

<

A

B

28 32003

where {.} denote the fractional part, is equal to

(A) 28 15 (B) 28 5 (C*) 28 19 (D) 28 9 Sol. 32003_{= 3}2001_.32 = 9(27)667 = 9(28 – 1)667 = 9[667C 28₀ 667₊ 1 667_{C (28)}666_{+ ... +} 667 667_C ₍ –1)667]

that means if we divide 32003_{by 28, remainder is 19}

Thus

_<=

<

>

?

<@

<

A

B

28 32003 = 28 19

26. The middle term in the expansion of

10 x 1 x 3 4 5 6 7 8 % is (A) 10_C 1 _x 1 (B*) 10_C 5 (C) 10_C 6 (D) 10_C 7x .

Sol. Middle term =

2 2 10 T % _{= T} 5 + 1 = 10_C 5x 10 – 5 5 x 1 3 4 5 6 7 8 =10_C 5.

27. If |x| < 1, then the coefficient of xn_{in the expansion of (1 + x + x}2_{+ x}3_...)2_is

(A)n (B) n – 1 (C) n + 2 (D*) n + 1 Sol. (1 + x + x2_{+ ...)}2₌ 2 x – 1 1 ₃ 4 5 6 7 8 = (1 – x) –2 & T_{r + 1} = ( –2)( –3)...(_r_! –2 –(r –1)) ( –x)r = ! r ) 1 )(_– 1 r ...( 3 . 2 ) 1 (_– r % r xr = ! r ) 1 r ( r ... 3 . 2 . 1 ) 1 ( – 2r % xr_{= (r + 1) x}r & _{coefficient of x}r_{= r + 1} & _{coefficient of x}n_{= n + 1} 28. If a_2m =

/

0 m 2 0 r 2mCr 1 , then

/

0 m 2 0 r 2mCr r equals (A) (2m – 1) a_2m (B) 2ma_2m (C*) ma_2m (D) am 2 m Sol. E =

/

0 m 2 0 r 2mCr r =

/

0 m 2 0 r 2mCr r – m 2 $ _{2E =}

/

0

%

m 2 0 r 2mCr r – m 2 r = 2m

/

0 m 2 0 r 2mCr 1

29. Last three digits of the number N = 7100

– 3100 are (A) 100 (B) 300 (C) 500 (D*) 000 Sol. N = 7100_–₃100 = (10 – 3)100 – 3100 = (100C₀. 10100 – ... – 100C₉₉. 10.399) + 3100 – 3100 = (1000)N + 3100_–₃100 = (1000) N last 3 digits = 000

30. If the lines represented by x2

– 2pxy – y2 = 0 are rotated about origin through an angle C, one in clockwise

direction and other in anticlockwise direction, then the equation of the bisectors of the angle between the lines in new position is

(A) px2_{+ 2xy + py}2_{= 0} _(B) px2

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Page # 6 30. If the lines represented by x2_–_2pxy_–_y2_{= 0 are rotated about origin through an angle} C_{, one in clockwise}

direction and other in anticlockwise direction, then the equation of the bisectors of the angle between the lines in new position is

(A) px2_{+ 2xy + py}2_{= 0} _(B) px2

– 2xy + py2 = 0 (C) px2 – 2pxy – py2 = 0 (D*) px2 + 2xy – py2 = 0

Sol. Bisectors of x2_–_2pxy_–_y2_{= 0 is}

2 y – x2 2 = p – xy $ _px2_{+ 2xy}_–_py2_{= 0}

Lines in new position will also have same angle bisectors.

31. If 3a + 2b + 6c = 0, then the family of straight lines ax + by + c = 0 passes through a fixed point whose coordinates are given by

(A*) 6₇8 ₄ 53 3 1 , 2 1 (B) (2, 3) (C) (3, 2) (D) 6₇8 ₄ 53 2 1 , 3 1 Sol. ax + by + 6₇8 ₄ 53 6 b 2 – a 3 – = 0 6ax + 6by – 3a – 2b = 0

a(6x – 3) + b(6y – 2) = 0

x = 1/2, y = 1/3

32. The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is

(A) 3/2 (B*) 3/10 (C) 6 (D) none of these

Sol. Distance = 2 2 4 3 2 15 – 9 % = 10 3

33. A ray of light passing through the point A (1, 2) is reflected at a point B on the x-axis and then passes through C (5, 3) . Then the equation of AB

is-(A*) 5x + 4y = 13 (B) 5x – 4y = – 3 (C) 4x + 5y = 14 (D) 4x – 5y = – 6

Sol.

AB will pass through CD.

& _{equation of AB is} _{y + 3 =} 4 – 5 (x – 5) $ 4y + 5x = 13

34. If bx + cy = a, where a, b, c are the same sign, be a line such that the area enclosed by the line and the axes of reference is

8 1

unit2_{, then}

(A) b, a, c are in G.P. (B) b, 2a, c are in A.P. (C) b,

2 a

, c are in A.P.. (D*) b, –2a, c are in G.P.

]Sol. bx + cy = a a b x + a c y = 1

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Page # 7 ]Sol. bx + cy = a a b x + a c y = 1 3 4 5 6 7 8 b a x + 3 4 5 6 7 8 c a y = 1 Area ofEOAB = 8 1 (given) $ 2 1 . b a . c a = ± 8 1 bc a2 = ± 4 1 $ _4a2_{= ± (bc)} $ _(2a)2_{= ± bc} b, ±2a, c are in G.P.

35. In a EABC if A is (1, 2) and equation of the medians through B and C are x + y = 5 and x = 4 respectively then B is

(A) (1, 4) (B*) (7, – 2) (C) (4, 1) (D) ( –2, 7)

Sol.

point of intersection of x + y = 5 and x = 4 G F (4, 1) & 3 4 b 1% % = 4 $ b = 7 So BF (7, – 2)

36. Let B₁ = 3x + 4y – 7 = 0 & B₂F 4x – 3y – 14 = 0 are angle bisectors of the angle between the lines L₁ = 0 &

L₂ = 0 in which L₁ is passes through the point (1, 2) then

(A*) B₁ is acute angle bisector (B) B₂ is acute angle bisector (C) B₁ & B₂ both are right angle bisector (D) Data is insufficient

Sol. Let d₁ & d₂ are the distance of point (1, 2) from the bisector B₁ & B₂. d₁ = 5 7 8 3% ; = 5 4 d₂ = 5 14 6 4; ; = 5 16 ! d₁ < d₂

& B₁ is an acute angle bisector

37. If P is a point (x, y) on the line y = – 3x such that P and the point (3, 4) are on the opposite sides of the

line 3x – 4y = 8, then

(A*) y < – 5 8 (B) y > – 5 8 (C) y > – 5 11 (D) y < – 5 1 Sol. Since 3.3 – 4.4 – 8 = – 15 < 0 $ _3x – 4y – 8 > 0 $ 3 3 4 5 6 7 8 3 y – – 4y – 8 > 0

(18)

Page # 8 Sol. Since 3.3 – 4.4 – 8 = – 15 < 0 $ 3x – 4y – 8 > 0 $ 3 ₄3 5 6 7 8 3 y – – 4y – 8 > 0 $ 5y < – 8$ y < – 5 8

38. If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are

in-(A) H.P. (B) G.P. (C*) A.P. (D) None of these

Sol. Since the given lines are concurrent,

& 1 14 c 1 13 b 1 12 a = 0 $ 0 1 b – c 1 13 b 0 1 – b – a = 0 [Applying R₃G R₃ – R₂, R₁G R₁ – R₂] $ _a – b + c – b = 0or 2b = a + c $ a, b, c are in A.P.

39. A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is-(A) 3 1 (B) 3 2 (C)1 (D*) 3 4 Sol. Equation of the line through the point (2, 2)

andH to line (1) is (y – 2) =

3 1 (x – 2) $ _3y – 6 = x – 2 $ x – 3y + 4 = 0 Its y-intercept = 3 4 . [Putting x = 0]

40. The line segment joining the points (1, 2) and (k, 1) is divided by the line 3x + 4y – 7 = 0 in the ratio

4 : 9, then k is-(A*) – 2 (B) 2 (C) –3 (D) 3 Sol. L : 3x + 4y – 7 = 0 – L (1, 2) : L (k, 1) = 4 : 9 $ – (3 + 8 – 7) : (3k + 4 – 7) = 4 : 9 $ – 4 : (3k – 3) = 4 : 9 $ k = – 2. 41. If px4_{+ qx}3_{+ rx}2_{+ sx + t =} x 3 4 x 3 x 3 x x 2 1 x 3 x 1 x x 3 x2 % ; ; ; % % ; % then t is equal to (A)33 (B) 20 (C)15 (D*)21 Sol. px4_{+ qx}3_{+ rx}2_{+ sx + t =} x 3 4 x 3 – x 3 – x x – 2 1 x 3 x 1 – x x 3 x2 % % % % Putting x = 0 t = 0 4 3 – 3 – 2 1 3 1 – 0 = 21

42. If the system of linear equations x + 2ay + az = 0, x + 3by + bz = 0, and x + 4 cy + cz = 0 has a non-zero solution, then a, b, c

(A) are in AP (B) are in GP (C*) are in HP (D) satisfy a + 2b + 3c = 0

Sol. The system of linear equations has a non-zero solution, then

a a 2 1

(19)

Page # 9 Sol. The system of linear equations has a non-zero solution, then

c c 4 1 b b 3 1 a a 2 1 = 0 Applying R₂G R₂ – R₁, R₃G R₃ – R₁ a – c a 2 – c 4 0 a – b a 2 – b 3 0 a a 2 1 = 0

$ (3b – 2a) (c – a) – (4c – 2a) (b – a) = 0

$ 3bc – 3ba – 2ac + 2a2 = 4bc – 2ab – 4ac + 2a2

$ 4ac – 2ac = 4bc – 2ab – 3bc + 3ab

$ _{2ac = bc + ab}

On dividing by abc, we get

b 2 = a 1 + c 1 Hence, a, b, c are in HP.

43. If a, b, c are pth, qth and rth, terms of a G.P., then

1 r c log 1 q b log 1 p a log equals

-(A*)0 (B)1 (C)logabc (D)pqr

Sol. If A be the first term and R be the c.r. of G.P., then a = ARp –1_{, b=AR}q –1_{, c=AR}r –1

log a = logA + (p – 1)log R

& E = 1 r A log 1 q A log 1 p A log + 1 r R log ) 1 r ( 1 q R log ) 1 q ( 1 p R log ) 1 p ( ; ; ; = 0 + log R

1

1 r

1

1 q

1

1 p

;

=0 [byC₂ – C₁]

44. For positive numbers x, y, z, the numerical value of the determinant

1 y log x log z log 1 x log z log y log 1 z z y y x x is

(A*) 0 (B) 1 (C) 2 (D) None of these

Sol. Value of determinant

1 y log x log z log 1 x log z log y log 1 z z y y x x

= _log1_x._log1_y._log1_z

z log y log x log z log y log x log z log y log x log = 0

45. The number of values of ' r ' satisfying the equation, ₂ r 39 1 r 3 39

_C

_;

_C

; =39

C

_r2_;₁

;

39

C

3r is (A) 1 (B*) 2 (C) 3 (D) 4 Sol. 39C3r;1;39 C_r2 039 C_r2_;₁;39 C3r $ r 3 39 1 r 3 39_C

_%

_C ; = 39C_r2_;₁%39C_r2 r 2 40_{C =} 2 r 40 C

(20)

Page # 10 Sol. 39C3r;1;39 C_r2 039 C_r2_;₁;39 C3r $ r 3 39 1 r 3 39_C

_%

_C ; = 39C_r2_;₁%39C_r2 r 2 40_{C =} 2 r 40_C r2 _{= 3r or} _{r = 0, 3} or r2_{+ 3r = 40} $ _{r = 5,} –8 BITSAT(XII)_PT-2_Pg.No # 1

BITSAT

_–

XII/XIII

PT

_–

02

(21)

BITSAT(XII)_PT-2_Pg.No # 1

BITSAT

_–

XII/XIII

PT

_–

02

1. The average velocity of molecules of a gas of molecular weight M at temperature T is:

(A*)0 (B) M RT 3 (C) 8 RT M

!

(D) 2R T M

Sol. Average velocity of a molecule at any temperature is zero because of its random mo tion.

2. The ratio of r.m.s. speed to the r.ms. angular speed of a diatomic gas at certain temperature is: (assume m = mass of one molecule, M = molecular mass,

"

= moment of inertia of the molecules)

(A) 3 2 (B) 3 2 I M (C*) 3 2 I m (D) 1 Sol. kT 2 3 mV 2 1 2 _# kT 2 2 2 1_"_$₂ _# m 2 3 V _# " $

3. A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:

(A) 4 R T (B)5R T (C)15R T (D*) 11 R T

Sol. In an ideas gas internal energy = 2 f nRT U = 2 5

× RT + 4 ×

2 3 RT = 11 RT.T.

4. Maxwell’

s velocity distribution curve is given for the same quantity at two different temperatures. For

the given curves.

(A) T₁ > T₂ (B*) T₁ < T₂ (C) T₁

%

T₂ (D) T₁ = T₂ Sol. Higher is the temperature greate r is the most probable veloc ity.

5. In a process the density of a gas remains co nstant. If the temperature is doubled, then the change in the pressure will be:

(A*) 100 % Increase (B) 200 % Increase (C) 50 % Decrease (D) 25 % Decrease Sol. We have

&

=

RT PM 1 1 RT M P = 2 2 RT M P 1 2 1 1 T 2 P T P _# P₂ = 2P₁

6. 12 gm He and 4 gm H₂ is filled in a container of volume 20 litre maintained at temperature 300 K. The pressure of the mixture is nearly :

(A) 3 atm (B) 5 atm (C*) 6.25 atm (D) 12.5 atm

BITSAT(XII)_PT-2_Pg.No # 2 Sol. PV = n RT 300 31 8 4 12 ) + , .

(22)

BITSAT(XII)_PT-2_Pg.No # 2 Sol. PV = n RT P = V nRT = 3 10 20 300 31 8 2 4 4 12 ' ( ( ( ) * + , - . / . = 6.25

× 10

5_Pa

7. In an experiment the speeds of any five molecules of an ideal gas are recorded. The experiment is repeated N times where N is very large. The aver age of recorded values, is :

(A*) M RT 2 (B) M RT 8 ! (C) M RT 3 (D) M RT

Sol. When speed of 5 molec ules which are selected random ly, then the average is most likely to be equal to the most probable speed.

0

The average of these values is most likely equal to M RT 2

.

8. P-V diagram o f a cyclic process A

1

B

1

C

1

A is shown in figure. The temperature of the gas will be maximum at :

(A)A (B)B

(C*) a point between A and B (D) a point between B and C

Sol. Temperature at points A and B are equal. A to B temperature first in creases then decrease.

9. On an X temperature scale, water freezes at –

125.0° X and boils at 375.0° X. On a Y temperature

scale, water freezes at –

70.0°Y and boils at

–

30.0°Y. The value of temperature on X-scale equal to the

temperature of 50.0

°Y on Y-scale is :

&

(A) 455.0

° X

(B) –

125.0° X

(C*) 1375.0

° X

(D) 1500.0

° X

Sol. 500 ) 125 ( X

'

= 40 ) 70 ( Y

'

For Y = 50 X = 1375.0

°X

10. The amount of heat supplied to decrease the volume of an ice water mixture by 1 cm3_{without any}

change in temperature, is equal to : (

&

_ice = 0.9,

&

_water = 80 cal/gm)

(A) 360 cal (B) 500 cal (C*) 720 cal (D) None

Sol. x gm ice convert into x gm water

9 . 0 x –

x = 1

2

x = 1 . 0 9 . 0 = 9

0

Q = 9

× 80 = 720 cal

11. n moles of a gas filled in a container at temperature T is in equilibrium initially. If the gas is compressed slowly and isothermally to half its initial volume, the work done by the atmosphere on the piston is:

BITSAT(XII)_PT-2_Pg.No # 3 (A*) n R T 2 (B)

'

n R T 2 (C) n R T !_{n 2} 1 2

'

.

-

,

_*)

+

(D)

'

n R T l n 2

(23)

BITSAT(XII)_PT-2_Pg.No # 3 (A*) n R T 2 (B)

'

n R T 2 (C) n R T !_{n 2} 1 2

'

.

-

,

_*)

+

(D)

'

n R T l n 2 Sol. Work done by atmosphere = P_atm

3

V

= P_atm

2 V

...(i)

As ; Initially gas in container is in thermodynamic equilibrium with its surroundings.

0

Pressure inside cylinder = P_atm & PV = nRT

2

P_atmV = nRT or V = atm P nRT Putting in (1), W = 2 nRT

12. In the figure shown the pressure of the gas in state B is:

(A) 25 63 P₀ (B*) 25 73 P₀ (C) 25 48 P₀ (D) none of these Sol. AN = 3v₀ cos2₃₇

_º

P_B = _*) + , - . ( / 25 16 v 3 v v P 0 0 0 0 = _*) + , - . / 25 48 1 = P₀(73/25) Ans. (B)

13. A vessel contains an ideal monoatomi c gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion is:

(A)Q (B) 3 5 Q (C*) 2 5 Q (D) 2 3 Q BITSAT(XII)_PT-2_Pg.No # 4

.Sol. For process at constant pressure Q = nC_p

3

T = 2 5 nR

3

T and W= P

3

V = nR

3

T = 5 2 Q

(24)

.Sol. For process at constant pressure Q = nC_p

3

T = 2 5 nR

3

T and W= P

3

V = nR

3

T = 5 2 Q

14. A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process ABCA will be :

(A) 25% (B) 12.5% (C) 50% (D*) 13 100 % Sol. W = 2 1 P₀V₀ = 2 1 RT₀ . Heat absorbed = Q_AB + Q_BC = C_VT₀ + C_P2T₀ = 2 13 RT₀

0

Efficiency = 0 0 0 0 V P 2 13 V P 2 1

× 100

,_-. 0 0 # RT0_* +) 2 13 V P 2 13 " = 13 1

× 100 = 7.7 %

Ans.

15. 1 mole of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. Graph shows the volume V versus Q. The gas temperature is nearly equal to : (use R = 8.31 J/K.mole)

(A) 208.4 K (B) 268.2 K (C*) 312.6 K (D) 353.8 K Sol. For isothermal process

Q = nRT !_n 1 2 v v 1800 = 1

× 8.3 T

!_{n z} get T = 312.6 K

16. Curve in the figure shows an adiabatic compression of an ideal gas from 15 m3_{to 12 m}3_{, followed by an}

isothermal compression to a final volume of 3.0 m3_{. There are 2.0 moles of the gas. Total heat supplied to}

the gas is equal to : (!_{n2 = 0.693)}

(25)

BITSAT(XII)_PT-2_Pg.No # 5 3 12 15 400 p(Pa) V(m )3 (A) 4521 J (B) –

4521 J

(C*) –

6653 J

(D) –

8476 J

Sol. There is no heat transfer in adiabatic compression. In isothermal process

pwafd :)ks"e laihMu esa dksbZ m"ek LFkkukUrj.k ugha gksrk gSA lerkih izfØ;k esa

Q =W =P₁V₁ ln 1 2 V V = 400 x 12 ln 4 1 = –

6653 J

17. Two bodies A and B have emissi vities 0.5 and 0.8 respec tively. At some temperatures the two bodies have maximum spectral emissive powers at wavelength 8000

Å and 4000 Å respectively. The ratio of

their emissive powers at these temperatures is:

(A*) 128 5 (B)10 (C) 16 5 (D) None of these

Sol. Let the body have temperatures T₁ and T₂ respectively at wavelength

4

₁ = 8000

Å and

4

₂= 4000

Å.

0

From Wien’

s displacement law

4

T = constant

2

4

₁T₁=

4

₂T₂ or 8000

× T

₁ = 4000T₂ or 2 1 T T = 2 1 Emissive power = e

5

AT4

0

Ratio of emissive powers at these temperature is

= ₄ 2 2 4 1 1 T e T e = 8 . 0 5 . 0

×

4 2 1₎ * + , - . = 128 5

18. N(< 100) molecules of a gas have velocities 1, 2, 3... N km/s respectively. Then (A) rms speed and average speed of molecules is same.

(B) ratio of rms speed to average speed is

7

(2N + 1)(N + 1)/6N (C) ratio of rms speed to average speed is

7

(2N + 1)(N + 1)/6 (D*) ratio of rms speed to average speed of molecules is

) 1 N ( 6 ) 1 N 2 ( 2

/

_/

Sol. V_rms = N V ... V V₁2/ ₂2/ _N2 = N N ... 2 12/ 2/ / 2 = N(N 1₆)_N(2N 1) / /

2

V_rms = 6 ) 1 N 2 ( ) 1 N ( / / V_avg = N V ... V V₁/ ₂/ / _N = N N ... 2 1/ / / = N(₂N_N/1) = N₂/1 avg rms V V = 2 ₆(2₍_NN_//₁1₎) BITSAT(XII)_PT-2_Pg.No # 6

19. A solid spherical black body of radius r and uniform mass distribution is in free space. It emits power

‘

P

’

and its rate of colling is R then

(26)

19. A solid spherical black body of radius r and uniform mass distribution is in free space. It emits power

‘

P

’

and its rate of colling is R then

(A) R P

8

r2 _{(B*) R P}

₈

_r _(C)_R_P

₈

_1/r2 _{(D) R P}

₈

r 1 Sol. Rate of radiation per unit area is proportional to (T4₎

0

P

8

AT4

2

P

8

r2_. Also ms dt dT

9

ATT4

0

dt dT = R

9

r 1

(because m = (v

&

)

9

r3_{and A}

₉

_r2₎

20. A black body emits radiati on at the rate P when its abso lute temperature is T. At this temperature the wavelength at which the radiation has maximum spectral emissive power is

4

₀. If at another temperature T

:

the power radiated is P

:

and wavelength at maximum spectral emissive power is

2

0

4

then

(A*) P

:

T

:

= 32PT (B) P

:

T

:

= 16PT (C) P

:

T

:

= 8PT (D) P

:

T

:

= 4PT Sol. For a black body, wavelength for maximum intensity :

T 1 48 _& _P

₈

_T4

2

P 1₄ 4 8

₂

_P

_:

_{= 16 P.} _P

_:

_T

_:

_{= 32PT}

21. Thermal coefficient of volume expansion at constant pressure for an ideal gas sample of n moles having pressure P₀, volume V₀ and temperature T₀ is

(A) 0 0V P R (B) R V P₀ ₀ (C*) 0 T 1 (D) 0 T n 1 Sol. [Easy] PV = nRT PdV = nRdT

;

= dT dV V 1 and P nR dT dV

_#

;

= T 1

For given temperature T₀,

;

=

0

T 1

22. A solid sphere of iron at 2

°C is lying at the bottom of a bucket ful l of water at 2°C. If the temperature of

the water is increased to 3

°C, the buoyant force on the sphere due to water will

(A*) Increase (B) Be unchanged (C) Decrease

(D) Increase or decrease depends upon the nume rical values of coeffici ent of expansion of water and iron.

Sol. As the temperature of water is increased from 2

°C to 3°C the density of water increases (remember

anamolous behaviour of water), also the volume of sphere increases. Therefore bouyant force on sphere due to water shall increase.

23. The lengths of two metallic rods at temperatures

<

are L_A and L_B and their linear coefficient of expansion are

8

_A and

8

_B respectively. If the difference in their lengths is to remain constant at any temperature then (A) L_A /L_B=

8

_A /

8

_B (B*) L_A /L_B=

8

_B /

8

_A (C)

8

_A=

8

_B (D)

8

_A

8

_B=1

Sol. Change in L_A = change in L_B i.e.

3

L_A =

3

L_B

2

8

_A

3

T L_A =

8

_B

3

TL_B or

8

_AL_A =

8

_BL_B .

24. Two identical long, solid cylinders are used to conduct heat from temp T₁ to temp T₂. Originally the cylinder are connected in series and the rate of heat transfer is H. If the cylinders are connected in parallel then the rate of heat transfer would be :