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Molarity

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Question

Calculate the osmotic pressure (in atm) of a solution having $0.065 M K C l, 0.02 M B a C l_{2} a n d 0.05 M H A$ (a weak acid which is 20% dissociated) at 300 Kelvin. $[U s e R = 0.08 l i t - a t m m o l^{- 1} K^{- 1}]$

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Solution

$p = (i_{1} C_{1} + i_{2} C_{2} + i_{3} C_{3}) R T$
$= (2 \times 0.065 + 3 \times 0.02 + 1.2 \times 0.05) \times 0.08 \times 300$
$= (0.13 + 0.06 + 0.06) \times 0.08 \times 300 = 6 a t m .$

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