The weight in grams of a non-volatile solute | 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_A}{n_A + n_B} = \frac{w_A/m_A}{w_A/m_

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{n_A}{n_A + n_B} = \frac{w_A/m_A}{w_A/m_A + w_B/m_B}$.Here,$w_A$ 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 water $(H_2O)$.Given: $m_A = 60$,$w_B = 90 \ g$,$m_B = 18 \ g/mol$,and relative lowering $= 0.02$.Substituting the values: $0.02 = \frac{x/60}{x/60 + 90/18}$.$0.02 = \frac{x/60}{x/60 + 5}$.$0.02(x/60 + 5) = x/60$.$0.02(x/60) + 0.1 = x/60$.$0.1 = x/60(1 - 0.02) = x/60(0.98)$.$x = \frac{0.1 	imes 60}{0.98} = \frac{6}{0.98} \approx 6.12 \ g$. Given the options,the closest integer value is $6 \ g$.

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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