At a given temperature,the vapour pressure of | vedclass

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(C) The relative lowering of vapour pressure is given by: $\frac{p^0 - p}{p^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (for dilute solutions)

Vedclass · Accepted Answer

$1.11$

Vedclass · Answer

(C) The relative lowering of vapour pressure is given by: $\frac{p^0 - p}{p^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (for dilute solutions).Here,$p^0 = 25 \ mm$,$p = 24.5 \ mm$.$\frac{25 - 24.5}{25} = \frac{0.5}{25} = 0.02$.For a dilute aqueous solution,the relationship between relative lowering of vapour pressure and molality $(m)$ is: $\frac{p^0 - p}{p^0} = m 	imes \frac{M_{solvent}}{1000}$.Given $M_{water} = 18 \ g/mol$,we have $0.02 = m 	imes \frac{18}{1000}$.$m = \frac{0.02 	imes 1000}{18} = \frac{20}{18} = 1.11 \ m$.

At a given temperature,the vapour pressure of pure water is $25 \ mm$ and the vapour pressure of a dilute solution of urea is $24.5 \ mm$. The molality of the solution is:

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