38.4 g of unknown substance (molar mass 384 | vedclass

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(C) Step $1$: Calculate the number of moles of solute $(n_2)$ and solvent $(n_1)$.$n_2 = \frac{38.4 \ g}{384 \ g \ mol^{-1}} = 0.1 \ mol$$n_1 = \frac{

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$0.7999$

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(C) Step $1$: Calculate the number of moles of solute $(n_2)$ and solvent $(n_1)$.$n_2 = \frac{38.4 \ g}{384 \ g \ mol^{-1}} = 0.1 \ mol$$n_1 = \frac{116 \ g}{58 \ g \ mol^{-1}} = 2 \ mol$Step $2$: Calculate the mole fraction of the solvent $(x_1)$.$x_1 = \frac{n_1}{n_1 + n_2} = \frac{2}{2 + 0.1} = \frac{2}{2.1} \approx 0.9524$Step $3$: Apply Raoult's law to find the vapour pressure of the solution $(P)$.$P = x_1 	imes P_0 = 0.9524 	imes 0.842 \ atm \approx 0.8019 \ atm$.Note: Based on the provided options,the closest value is $0.7999 \ atm$.

$38.4 \ g$ of unknown substance (molar mass $384 \ g \ mol^{-1}$) and $116 \ g$ of acetone is used to prepare a solution at $313 \ K$. If vapour pressure of pure acetone (molar mass $58 \ g \ mol^{-1}$) is $0.842 \ atm$,what is the vapour pressure of the solution (in $atm$)?

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