A solution of a substance having mass $1.8 \times 10^{- 3} k g$ has the osmotic pressure of $0.52 a t m$ at $280 K$ . Calculate the molar mass of the substance used.
[ $V o l u m e = 1 d m^{3}, R = 0.0821 L a t m K^{- 1} m o l^{- 1}]$ .

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Solution

osmotic pressure $= 0.52 a t m$
$π = 0.52 a t m$
$π = C R T$
$0.52 C \times 0.0821 \times 280$
$C = \frac{0.52}{280 \times 0.0821} = 0.02262$
mass of solute $= 1.8 \times 10^{- 3} k g = 1.8 g m$
Molarity $= \frac{1.8}{m} / 1.4 l r e$
$\frac{1.8}{M} = 0.02262$
$M = \frac{1.8}{0.02262} = 79.57 \approx 80$

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