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Van't Hoff Factor

A 0.004 M sol...

Question

A $0.004 M$ solution of $N a_{2} S O_{4}$ is isotonic with a $0.010 M$ solution of glucose at same temperature. The apparent precentage degree of dissociation of $N a_{2} S O_{4}$ is:

25

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50

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75

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85

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Solution

The correct option is C $75$ %
Isotonic solution:
Two solutions having the same osmotic pressure across a semipermeable membrane is referred to as an isotonic solution

Osmotic pressure,
$π (N a_{2} S O_{4}) = i C R T = (0.004) i R T$
$π (g l u c o s e) = C R T = 0.010 R T ∵ i = 1 for non-electrolyte$
As solutions are isotonic,
$i (0.004) R T = 0.01 R T$
$i = 2.5$

$N a_{2} S O_{4} ⇌ 2 N a^{+} + S O_{4}^{2 -}$

1 mol 0 0

$1 - α$ $2 α$ $α$ ,

$n = 3$
we know,
$i = 1 + α (n - 1)$
$i = 1 + 2 α$

$α = \frac{i - 1}{2} = \frac{2.5 - 1}{2} = \frac{1.5}{2} = 0.75 = 75 %$

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Van't Hoff Factor

Standard XII Chemistry

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