The vapour pressure of solution of 5g of non electrolyte in 100 g of a water at a particular temperature is $2985 N m^{- 2}$ . The vapour pressure of pure water at that temperature is $3000 N m^{- 2}$

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Solution

Mole fraction of solute, $x_{2} = \frac{- Δ p}{p_{1}^{*}} = \frac{3000 - 2985}{3000} = 0.005$
Now, $x_{2} = \frac{m_{2} / M_{2}}{\frac{m_{1}}{M_{1}} + \frac{m_{2}}{M_{2}}} \approx \frac{m_{2} / M_{2}}{m_{1} / M_{1}}$
$∴ 0.005 = \frac{5 / M_{2}}{100 / 18}$
$\Rightarrow M_{2} = 180 g m o l^{- 1}$
$∴$ The molar mass of solute is $180 g m o l^{- 1}$ .

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The vapour pressure of solution of 5g of non electrolyte in 100 g of a water at a particular temperature is 2985Nm−2. The vapour pressure of pure water at that temperature is 3000Nm−2

Mole fraction of solute, x2=−Δpp∗1=3000−29853000=0.005Now, x2=m2/M2m1M1+m2M2≈m2/M2m1/M1∴0.005=5/M2100/18⇒M2=180gmol−1∴ The molar mass of solute is 180gmol−1 .

The vapour pressure of solution of 5g of non electrolyte in 100 g of a water at a particular temperature is $2985 N m^{- 2}$ . The vapour pressure of pure water at that temperature is $3000 N m^{- 2}$