Find the molar mass of solute when 2 g is di | vedclass

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(C) The formula for relative lowering of vapour pressure is given by: $\frac{\Delta P}{P_{A}^{\circ}} = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A}

Vedclass · Accepted Answer

$43.3 \ g \ mol^{-1}$

Vedclass · Answer

(C) The formula for relative lowering of vapour pressure is given by: $\frac{\Delta P}{P_{A}^{\circ}} = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A} = \frac{W_B}{M_B} 	imes \frac{M_A}{W_A}$ Given: $\frac{\Delta P}{P_{A}^{\circ}} = 0.06$,$W_B = 2 \ g$,$W_A = 60 \ g$,$M_A = 78 \ g \ mol^{-1}$. Substituting the values: $0.06 = \frac{2}{M_B} 	imes \frac{78}{60}$ $0.06 = \frac{156}{60 	imes M_B}$ $M_B = \frac{156}{60 	imes 0.06} = \frac{156}{3.6} = 43.33 \ g \ mol^{-1}$ Thus,the molar mass of the solute is $43.3 \ g \ mol^{-1}$.

Find the molar mass of solute when $2 \ g$ is dissolved in $60 \ g$ of benzene,and the relative lowering of vapour pressure is $0.06$. (Molar mass of benzene is $78 \ g \ mol^{-1}$)

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