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JEE Advanced 2021 Paper 2 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

JEE Advanced 2021 Paper 2 Chemistry: Paragraph 2

A solute, benzoic acid (M=122 g mol−1M=122\ \mathrm{g\,mol^{-1}}), partially dimerises through hydrogen bonding in benzene according to 2A⇌A22\mathrm{A}\rightleftharpoons \mathrm{A_2}, and does not ionise. A solution of 3.00 g3.00\ \mathrm{g} of benzoic acid in 100 g100\ \mathrm{g} of benzene freezes 0.755 K0.755\ \mathrm{K} below pure benzene. For benzene Kf=5.12 K kg mol−1K_f=5.12\ \mathrm{K\,kg\,mol^{-1}} and the density at the experimental temperature is 0.877 g mL−10.877\ \mathrm{g\,mL^{-1}}. Treat the solution volume as equal to the volume of benzene, and let α\alpha be the fraction of the benzoic acid molecules that are associated into dimers, that is, the degree of association.

Question 1

Numerical answer

The degree of association α\alpha of benzoic acid in benzene under these conditions is ______. Give the answer correct to two decimal places.
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Answer: 0.80

The molality is m=(3.00/122)/0.100=0.2459 mol kg−1m=(3.00/122)/0.100=0.2459\ \mathrm{mol\,kg^{-1}}. From ΔTf=i Kf m\Delta T_f=i\,K_f\,m, the van't Hoff factor is i=0.755/(5.12×0.2459)=0.5997i=0.755/(5.12\times0.2459)=0.5997. For association 2A⇌A22\mathrm{A}\rightleftharpoons \mathrm{A_2} starting with 11 mole of monomer units, the moles at equilibrium are (1−α)+α/2=1−α/2(1-\alpha)+\alpha/2=1-\alpha/2, so i=1−α/2i=1-\alpha/2 (a value below 11, as expected when particle number decreases). Hence α=2(1−i)=2(1−0.5997)=0.80\alpha=2(1-i)=2(1-0.5997)=0.80. Ignoring that i<1i<1 and using i=1+αi=1+\alpha would give a negative degree of association, flagging the error. The fraction 0.800.80 means that eighty percent of the benzoic acid molecules are bound in dimers.

Question 2

Numerical answer

The equilibrium constant KcK_c (in L mol−1\mathrm{L\,mol^{-1}}) for 2A⇌A22\mathrm{A}\rightleftharpoons \mathrm{A_2} in benzene, taking α=0.80\alpha=0.80 exactly, is ______. Give the answer correct to two decimal places.
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Answer: 46.37

The volume of benzene is 100/0.877=114.0 mL=0.1140 L100/0.877=114.0\ \mathrm{mL}=0.1140\ \mathrm{L}, and the volume of solution is taken as equal to this. The total monomer-unit concentration is c0=(3.00/122)/0.1140=0.2157 mol L−1c_0=(3.00/122)/0.1140=0.2157\ \mathrm{mol\,L^{-1}}. With α=0.80\alpha=0.80, the monomer concentration is [A]=c0(1−α)=0.04313[\mathrm{A}]=c_0(1-\alpha)=0.04313 and the dimer concentration is [A2]=c0α/2=0.08626[\mathrm{A_2}]=c_0\alpha/2=0.08626 in mol L−1\mathrm{mol\,L^{-1}}. Hence Kc=[A2]/[A]2=0.08626/(0.04313)2=46.37 L mol−1K_c=[\mathrm{A_2}]/[\mathrm{A}]^2=0.08626/(0.04313)^2=46.37\ \mathrm{L\,mol^{-1}}. Note that the dimer count is α/2\alpha/2 of the monomer units, not α\alpha; forgetting the factor 12\tfrac12 would double the answer.
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