If 8 ; g of a non-electrolyte solute is disso | vedclass

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(B) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by:$\frac{P_{A}^{0} - P_{s}}{P_{A}^{0}} = \fr

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$40$

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(B) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by:$\frac{P_{A}^{0} - P_{s}}{P_{A}^{0}} = \frac{n_{B}}{n_{A} + n_{B}} \approx \frac{n_{B}}{n_{A}}$ (for dilute solutions).Given:$w_{B} = 8 \; g$,$W_{A} = 114 \; g$,$M_{A} = 114 \; g \; mol^{-1}$.Since the vapour pressure is reduced to $80 \%$,$P_{s} = 0.8 P_{A}^{0}$.Therefore,$\frac{P_{A}^{0} - 0.8 P_{A}^{0}}{P_{A}^{0}} = \frac{0.2 P_{A}^{0}}{P_{A}^{0}} = 0.2$.Using the formula $\frac{\Delta P}{P_{A}^{0}} = \frac{w_{B} 	imes M_{A}}{m_{B} 	imes W_{A}}$:$0.2 = \frac{8 	imes 114}{m_{B} 	imes 114}$.$0.2 = \frac{8}{m_{B}}$.$m_{B} = \frac{8}{0.2} = 40 \; g \; mol^{-1}$.

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

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If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.