3 g of urea is dissolved in 45 g of H_2O. T | vedclass

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(C) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{n_2}{n_1+n_2}$ where $p^{\circ}$ is the

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$0.02$

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(C) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{n_2}{n_1+n_2}$ where $p^{\circ}$ is the vapour pressure of the pure solvent,$p_s$ is the vapour pressure of the solution,$n_2$ is the number of moles of the solute (urea),and $n_1$ is the number of moles of the solvent $(H_2O)$. Molar mass of urea $(NH_2CONH_2)$ = $60 \ g/mol$. $n_2 = \frac{w_2}{M_2} = \frac{3 \ g}{60 \ g/mol} = 0.05 \ mol$. Molar mass of $H_2O = 18 \ g/mol$. $n_1 = \frac{w_1}{M_1} = \frac{45 \ g}{18 \ g/mol} = 2.5 \ mol$. Substituting the values: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{0.05}{2.5 + 0.05} = \frac{0.05}{2.55} \approx 0.0196 \approx 0.02$.

$3 \ g$ of urea is dissolved in $45 \ g$ of $H_2O$. The relative lowering in vapour pressure is

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