The vapour pressure of a solution containing 5 g of a non electrolyte in 100 g of water at a particular temperature is $2985 N m^{- 2}$ . If the vapour pressure of pure water is $3000 N m^{- 2}$ , the molecular weight of solute is:

60

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120

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180

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380

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Solution

The correct option is C $180$
we have formula of $Relative lowering of vapour pressure$ which is

$\frac{P_{o} - P}{P_{o}} = X_{s o l u t e} = \frac{n_{s o l u t e}}{n_{s o l u t e} + n_{s o l v e n t}} \approx \frac{n_{s o l u t e}}{n_{s o l v e n t}}$

Now, pute the values

$\frac{3000 - 2985}{3000} = \frac{\frac{5}{M}}{\frac{100}{18}}$

$\frac{15}{3000} = \frac{5 \times 18}{100 \times M}$

$M = 180$

correct answer is C.

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The vapour pressure of a solution containing 5 g of a non electrolyte in 100 g of water at a particular temperature is 2985Nm−2. If the vapour pressure of pure water is 3000Nm−2, the molecular weight of solute is:

The correct option is C 180we have formula of Relative lowering of vapour pressure which isPo−PPo=Xsolute=nsolutensolute+nsolvent≈nsolutensolventNow, pute the values3000−29853000=5M10018153000=5×18100×MM=180correct answer is C.

The vapour pressure of a solution containing 5 g of a non electrolyte in 100 g of water at a particular temperature is $2985 N m^{- 2}$ . If the vapour pressure of pure water is $3000 N m^{- 2}$ , the molecular weight of solute is: