The freezing point of a solution containing $0.2$ g of acetic acid in $20.0$ g of benzene is lowered by ${0.45}^{0}$ C. Calculate the degree of association of acetic acid in benzene. $K_{f}$ for benzene $= 5.12 K m o l^{- 1} k g$

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Solution

Depression of freezing point,
${Δ T}_{f} = K_{f} \times \frac{w t . o f {C H}_{3} C O O H \times 1000}{O b s e r v e d M o l . w t . o f {C H}_{3} C O O H \times w t . o f b e n z e n e}$

$\Rightarrow 0.45 = 5.12 \times \frac{0.2 \times 1000}{O b s e r v e d m o l . w t . o f {C H}_{3} C O O H \times 20}$

$\Rightarrow$ Observed Mol. wt. of ${C H}_{3} C O O H = 113.78$

$2 {C H}_{3} C O O H ⇌ {({C H}_{3} C O O H)}_{2}$
$1 - α$ $α / 2$

Where the degree of dissociation is $α$ .

Total no. of moles of particles at equilibrium

$i = 1 - α + α / 2 = 1 - α / 2$

now, $i = \frac{N o r m a l m o l e c u l a r w t . o f {C H}_{3} C O O H}{O b s e r v e d m o l e c u l a r w t . o f {C H}_{3} C O O H}$

$\Rightarrow 1 - α / 2 = \frac{60}{113.78} \Rightarrow α = 0.945$

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0.2

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