CBSE Class 12 Chemistry 2020 Outside Delhi Set 1 Solved Paper

Question : 28 of 37

Marks: +1, -0

SECTION -C

The freezing point of a solution containing

5 {\g}

of benzoic acid

(M=122 {\g} {\mol}^{-1})

35 {\g}

of benzene is depressed by

2.94 {\K}

. What is the percentage association of benzoic acid if it forms a dimer in solution?

( {\K}_f \;\text " for benzene "\;=4.9 {\K} {\kg} {\mol}^{-1})

Solution:

Observed molar mass of benzoic acid:

\;M_B=\;{K_f × W_B}/{W_A × ∆ T_f}

\;W_B=5 {\gm}

\;W_A=0.035 {\kg}

\;M_B=\;{4.9 × 5}/{0.035 × 2.94}=\;{24.5}/{0.1029}

\;=238 {\g} {\mol}^{-1}

\;K_f=4.9 {\K} {\kgmol}^{-1}

\;∆ T_f=2.94

\;\;\text " Normal molar mass of "\; {\C}_6 {\H}_5 {\COOH}

\;=122 {\g} {\mol}^{-1}

\;i=\;{\;\text " normal molar mass "\;}/{\;\text " observed molar mass "\;}=\;{122}/{238}=0.513

\;\% \;\text " of association of acid "\;(α)

\;2 {\C}_6 {\H}_5 {\COOH} ⇌ ( {\C}_6 {\H}_5 {\COOH}) _2

\;n=2 α=\;{i-1}/{1 \/ n-1}=\;{0.513-1}/{{1}/{2}-1}

=\;{(-0.487)}/{(-0.5)}=0.974

Percentage association of acid

=0.974 \×100=

97.4 \%

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