At 25,^oC,the vapour pressure of CCl_4 is 143 | vedclass

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(A) Mass of $CCl_4 = 	ext{Volume} 	imes 	ext{Density} = 100\,mL 	imes 1.58\,g/mL = 158\,g$. Molar mass of $CCl_4 (C=12, Cl=35.5) = 12 + 4 	imes 3

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$141.93$

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(A) Mass of $CCl_4 = 	ext{Volume} 	imes 	ext{Density} = 100\,mL 	imes 1.58\,g/mL = 158\,g$. Molar mass of $CCl_4 (C=12, Cl=35.5) = 12 + 4 	imes 35.5 = 154\,g/mol$. Moles of solvent $(N_{solvent})$ $= 158 / 154 \approx 1.026\,mol$. Moles of solute $(n_{solute})$ $= 0.5 / 65 \approx 0.00769\,mol$. Using Raoult's Law: $(P^0 - P_s) / P_s = n_{solute} / N_{solvent}$. $(143 - P_s) / P_s = (0.5 / 65) / (158 / 154)$. $(143 - P_s) / P_s = (0.5 	imes 154) / (65 	imes 158) = 77 / 10270 \approx 0.0075$. $143 - P_s = 0.0075 P_s \Rightarrow 143 = 1.0075 P_s$. $P_s = 143 / 1.0075 \approx 141.93\,mm\,Hg$.

At $25\,^oC$,the vapour pressure of $CCl_4$ is $143\,mm\,Hg$. If $0.5\,g$ of a non-volatile solute (molar mass $= 65\,g/mol$) is dissolved in $100\,mL$ of $CCl_4$,what will be the vapour pressure of the solution? (Density of $CCl_4 = 1.58\,g/cm^3$)

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