A 1.2% solution (w/v) of NaCl is isotonic with 7.2% solution (w/v) of glucose. Calculate degree of ionization and Van't Hoff factor of $N a C l$ .

0.55, 0.45

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0.95, 1.95

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0.98, 2.05

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0.95, 2.76

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Solution

The correct option is B $0.95, 1.95$
The molar masses of $N a C l$ and glucose are $58.5$ g/mol and $180$ g/mol respectively.
The expression for the osmotic pressure is as follows:
$Π = i C R T$
Since two solutions are isotonic, they have equal osmotic pressure.
$Π (N a C l) = Π (g l u c o s e)$
$i (N a C l) C (N a C l) = i (g l u c o s e) C (g l u c o s e)$
$i (N a C l) \times \frac{1.2}{58.5} = 1 \times \frac{7.2}{180} i (N a C l) = 1.95$
Let $α$ be the degree of ionization of $N a C l$ .
$α = \frac{i - 1}{n^{'} - 1} = \frac{1.95 - 1}{2 - 1} = 0.95$
Here, $n^{'}$ represents the total number of ions produced by dissociation of one molecule of $N a C l$ .

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A 1.2% solution (w/v) of NaCl is isotonic with 7.2% solution (w/v) of glucose. Calculate degree of ionization and Van't Hoff factor of NaCl.

A 1.2% solution (w/v) of NaCl is isotonic with 7.2% solution (w/v) of glucose. Calculate degree of ionization and Van't Hoff factor of $N a C l$ .