The weight in grams of a non-volatile solute | vedclass

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(C) The formula for relative lowering of vapour pressure is given by: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{n_A}{n_A+n_B} = \frac{\frac{w_A}{m_A}}{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The formula for relative lowering of vapour pressure is given by: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{n_A}{n_A+n_B} = \frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A}+\frac{w_B}{m_B}}$Where $w_A$ and $m_A$ are the mass and molar mass of the solute, and $w_B$ and $m_B$ are the mass and molar mass of the solvent (water).Given: $\frac{p^{\circ}-p_s}{p^{\circ}} = 0.02$, $m_A = 60 \ g/mol$, $w_B = 90 \ g$, $m_B = 18 \ g/mol$.Substituting the values: $0.02 = \frac{\frac{w_A}{60}}{\frac{w_A}{60} + \frac{90}{18}}$$0.02 = \frac{\frac{w_A}{60}}{\frac{w_A}{60} + 5}$$0.02 	imes (\frac{w_A}{60} + 5) = \frac{w_A}{60}$$0.02 	imes \frac{w_A}{60} + 0.1 = \frac{w_A}{60}$$0.1 = \frac{w_A}{60} - 0.02 	imes \frac{w_A}{60}$$0.1 = \frac{w_A}{60} 	imes (1 - 0.02) = \frac{w_A}{60} 	imes 0.98$$w_A = \frac{0.1 	imes 60}{0.98} = \frac{6}{0.98} \approx 6.12 \ g$. Since the closest option is $6 \ g$, the correct answer is $C$.

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

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