What will be the molar mass of a non-volatile | vedclass

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(D) According to Raoult's law for non-volatile solutes: $\frac{P^{0} - P}{P^{0}} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}} = \frac{W_{

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$117.0 \ g \ mol^{-1}$

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(D) According to Raoult's law for non-volatile solutes: $\frac{P^{0} - P}{P^{0}} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}} = \frac{W_{2} 	imes M_{1}}{M_{2} 	imes W_{1}}$Given: $P^{0} = 450 \ mm \ Hg$,$P = 400 \ mm \ Hg$,$W_{2} = 1.5 \ g$,$W_{1} = 30 \ g$,$M_{1} (C_{6}H_{6}) = (6 	imes 12) + (6 	imes 1) = 78 \ g \ mol^{-1}$.Rearranging the formula for $M_{2}$:$M_{2} = \frac{W_{2} 	imes M_{1} 	imes P^{0}}{W_{1} 	imes (P^{0} - P)}$$M_{2} = \frac{1.5 	imes 78 	imes 450}{30 	imes (450 - 400)}$$M_{2} = \frac{1.5 	imes 78 	imes 450}{30 	imes 50} = \frac{1.5 	imes 78 	imes 9}{30} = \frac{1053}{9} = 117.0 \ g \ mol^{-1}$.

What will be the molar mass of a non-volatile solute if the vapour pressure of pure benzene is $450 \ mm \ Hg$ and it decreases to $400 \ mm \ Hg$ when $1.5 \ g$ of the solute is added to $30 \ g$ of benzene? (Atomic mass: $C=12, H=1$)

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