Which of the following pair of solutions is isotonic?
(Assume all are 100% ionised)

0.15 M N a C l

and

0.1 M K_{2} S O_{4}

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0.2 M B a (N O_{3})_{2}

and

0.3 M N a_{2} S O_{4}

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0.2 M B a C l_{2}

and

0.2 M U r e a

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None of the above

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Solution

The correct option is A $0.15 M N a C l$ and $0.1 M K_{2} S O_{4}$
Isotonic solutions are solutions having same osmotic pressure.
Osmotic pressure of a solution can be given by,
$π = i C R T$
$π$ = Osmotic pressure
C = Concentration ( $m o l / L$ )
R = Gas constant ( $L a t m m o l^{- 1} K^{- 1}$ )
T = Temperature ( $K$ )
$i$ = van't Hoff factor

(a)
$N a C l \to N a^{+} + C l^{-}$
$π_{N a C l} = 2 \times 0.15 R T = 0.3 R T$
$K_{2} S O_{4} \to 2 K^{+} + S O_{4}^{2 -}$
$∴$ $i = 3$
$π_{K_{2} S O_{4}} = 3 \times 0.1 R T = 0.3 R T$
Since
$π_{K_{2} S O_{4}} = π_{N a C l}$
So, isotonic

(b)
$B a (N O_{3})_{2} \to B a^{2 -} + 2 N O_{3}^{-}$
$∴$ $i = 3$
$π_{B a (N O_{3})_{2}} = 3 \times 0.2 R T$
$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{2 -}$
$∴$ $i = 3$
$π_{N a_{2} S O_{4}} = 3 \times 0.3 R T$
$π_{N a_{2} S O_{4}} \neq π_{B a (N O_{3})_{2}}$
So, not isotonic

(c)
$B a C l_{2} \to B a^{2 +} + 2 C l^{-}$
$∴$ $i = 3$
$π_{B a C l_{2}} = 3 \times 0.2 R T$
Urea is a non-electrolyte so, $i = 1$
$π_{U r e a} = 0.2 R T$
Since ,
$π_{B a C l_{2}} \neq π_{N a C l}$
So, not isotonic.

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