At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ______ g . (Given: R = 0.083 L bar ) [2023]

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Solution

Q.
At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ______ g ${mol}^{- 1}$ .

(Given: R = 0.083 L bar $K^{- 1}$ ${mol}^{- 1}$ ) [2023]

Ans.
(62250)

$π = C R T$

$\frac{400 \times 10^{- 5}}{(1.01325)} = [\frac{2.5 \times 1000}{M_{solute} \times 250}] (0.083 \times 300)$

$400 \times 10^{- 5} = \frac{10}{M_{solute}} \times 0.083 \times 300$

$M_{solute} = \frac{10 \times 0.083 \times 300}{400} \times 10^{5}$

$= [\frac{10 \times 0.083 \times 3}{4}] \times 10^{5} = 62250 gram/mole$

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