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(A) The relative lowering of vapour pressure is given by the mole fraction of the solute,which is $\frac{n_2}{n_1 + n_2}$.Given:Mass of urea $(w_2)$ $

Vedclass · Accepted Answer

$0.018$

Vedclass · Answer

(A) The relative lowering of vapour pressure is given by the mole fraction of the solute,which is $\frac{n_2}{n_1 + n_2}$.Given:Mass of urea $(w_2)$ $= 3 \ g$Molar mass of urea $(M_2)$ $= 60 \ g \ mol^{-1}$Moles of urea $(n_2)$ $= \frac{3}{60} = 0.05 \ mol$Mass of water $(w_1)$ $= 50 \ g$Molar mass of water $(M_1)$ $= 18 \ g \ mol^{-1}$Moles of water $(n_1)$ $= \frac{50}{18} \approx 2.778 \ mol$Relative lowering of vapour pressure $= \frac{n_2}{n_1 + n_2} = \frac{0.05}{2.778 + 0.05} = \frac{0.05}{2.828} \approx 0.01768 \approx 0.018$.

Calculate the relative lowering of vapour pressure of a solution containing $3 \ g$ urea in $50 \ g$ water. [ Molar mass of urea $= 60 \ g \ mol^{-1} ]$

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