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Q.

Calculate the percentage degree of dissociation of an electrolyte $X Y_{2}$ (Normal molar mass =164) in water. If the observed molar mass by measuring elevation in boiling point is 65.6

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a

75%

b

25%

c

65%

d

37.5 %

answer is A.

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Detailed Solution

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$A B_{2} \overset{}{\to} A + 2 B$

i = Vant Hoff factor = 1 + 2d (d = degree of dissociation)

$i = \frac{Normal mass}{Observed mass} = \frac{164}{65 .6} = 2 .5$

$\Rightarrow 1 + 2 d = 2 .5 \Rightarrow d = 0 .75,$ Thus percentage dissociation = 0.75 × 100 = 75%

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