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Standard XII

Chemistry

Elevation in Boiling Point

One molal sol...

Question

One molal solution of a carboxylic acid in benzene shows the elevation boliling point of 1.518 K. The degree of association for dimerization of the acid in benzene is $(k_{b}$ for benzene= 2.53 K kg $m o l^{-} 1)$

60%

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75%

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70%

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80%

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Solution

The correct option is C 80%
$Δ T_{b} (n o r m a l) = K_{b} m = 2.53 \times 1 = 2.53 K$
$i = \frac{Δ T_{b (o b s)}}{Δ T_{b (n o r)}} = \frac{1.518}{2.53} = 0.6$
$i = 1 - (1 - \frac{1}{n}) α; α = 0.8$

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One molal solution of a carboxylic acid in benzene shows the elevation boliling point of 1.518 K. The degree of association for dimerization of the acid in benzene is (kb for benzene= 2.53 K kg mol−1)

The correct option is C 80%ΔTb(normal)=Kbm=2.53×1=2.53Ki=ΔTb(obs)ΔTb(nor)=1.5182.53=0.6i=1−(1−1n)α;α=0.8

One molal solution of a carboxylic acid in benzene shows the elevation boliling point of 1.518 K. The degree of association for dimerization of the acid in benzene is $(k_{b}$ for benzene= 2.53 K kg $m o l^{-} 1)$

The correct option is C 80%
$Δ T_{b} (n o r m a l) = K_{b} m = 2.53 \times 1 = 2.53 K$
$i = \frac{Δ T_{b (o b s)}}{Δ T_{b (n o r)}} = \frac{1.518}{2.53} = 0.6$
$i = 1 - (1 - \frac{1}{n}) α; α = 0.8$