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AIIMS 2016
Exam
AIIMS
Year
2016
Subject
Chemistry
Chapter
Solutions

AIIMS 2016 · Chemistry · Solutions

The freezing point of a solution containing 0.2 ~g of acetic acid...

The freezing point of a solution containing 0.2 g0.2 \mathrm{~g} of acetic acid in 20.0 g20.0 \mathrm{~g} benzene is lowered by 0.45∘C0.45^{\circ} \mathrm{C}. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and KfK_f for benzene =5.12 K kg mol−1=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ) Mobserved M_{\text{observed }} of acetic acid =113.78=113.78

Options

  1. A94.5%94.5 \%
  2. B54.9%54.9 \%
  3. C78.2%78.2 \%
  4. D100%100 \%

Correct answer

A. 94.5%94.5 \%

Step-by-step solution

Given : w2=0.2 g,w1=20 g,ΔTf=0.45∘Cw_2=0.2 \mathrm{~g}, w_1=20 \mathrm{~g}, \Delta T_f=0.45^{\circ} \mathrm{C} ΔTf=1000×Kf×w2w1×M⇒0.45=1000×5.12×0.220×M\Delta T_f=\frac{1000 \times K_f \times w_2}{w_1 \times M} \Rightarrow 0.45=\frac{1000 \times 5.12 \times 0.2}{20 \times M} ∴M(observed) =113.78\therefore \quad M_{\text{(observed) }}=113.78 (acetic acid) 2CH3COOH⇌2 \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons\left(\mathrm{CH}_3 \mathrm{COOH}\right)_2BeforeassociationAfterassociationBefore association After association1-\alpha \alpha / 2(where(where\alphaisdegreeofassociation)Molecularweightofaceticacidis degree of association) Molecular weight of acetic acid=60 i= Normal molecular mass  Observed molecular mass ∴M(normal )M(observed) =1−α+α2 or, 60113.78=1−α+α2∴α=0.945 or 94.5%\begin{aligned} & \quad i=\frac{\text{ Normal molecular mass }}{\text{ Observed molecular mass }} \\ & \therefore \quad \frac{M_{(\text{normal })}}{M_{(\text{observed) }}}=1-\alpha+\frac{\alpha}{2} \\ & \text{ or, } \frac{60}{113.78}=1-\alpha+\frac{\alpha}{2} \therefore \alpha=0.945 \text{ or } 94.5 \%\end{aligned}
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