Calculate the amount of $C a C l_{2}$ (van't Hoff factor $i = 2.47$ ) dissolved in $2.5 L$ solution so that its osmotic pressure at $300 K$ is $0.75$ atmosphere.
Given: Molar mass of $C a C l_{2}$ is $111 g . m o l^{- 1}$ ,
$R = 0.082 L . a t m . K^{- 1} m o l^{- 1}$

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Solution

The osmotic pressure (calculate), $π = \frac{n}{V} R T . . . . (i)$
where, $V = v o l u m e = 2.5 L, T = T e m p e r a t u r e = 300 K, n =$ dissolving moles of
$C a C l_{2} = \frac{wt of solute (g)}{molecular mass of solute (M)}$
Thus equation (i) becomes $π_{C a l} = \frac{8}{M V} R T$
$∴ π_{C a l} = \frac{g}{111 g m o l^{- 1} \times 2.5 l} \times 0.082 l a t m K^{- 1} m o l^{- 1} \times 300 K$
$= \frac{24.6 g}{277.5} a t m = 0.0886 \times g . a t m . g m$

Van't Hoff factor (i) $= \frac{observed osmotic pressure (π_{o b})}{calculated osmotic pressure (π_{c a l})}$
$∴ 2.47 = \frac{0.75 a t m}{π_{C a l}}$ or, $π_{C a l} = 0.3036 a t m$
Thus, $0.3036 = 0.0886 \times g g m$ or,
$g = 3.4271 g m$ .
Hence, the amount of $C a C l_{2}$ dissolve in $2.71$ solution is $3.4271 g m$ .

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