At 25^circ C,the vapor pressure of CCl_4 is 1 | vedclass

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(A) Given: Density of $CCl_4$ $(d)$ $= 1.58 \ g/cm^3$,Volume of $CCl_4$ $(V)$ $= 100 \ mL$.Mass of $CCl_4$ $(W_0)$ $= d 	imes V = 1.58 	imes 100 = 1

Vedclass · Accepted Answer

$141.93$

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(A) Given: Density of $CCl_4$ $(d)$ $= 1.58 \ g/cm^3$,Volume of $CCl_4$ $(V)$ $= 100 \ mL$.Mass of $CCl_4$ $(W_0)$ $= d 	imes V = 1.58 	imes 100 = 158 \ g$.Molar mass of $CCl_4$ $(M_0)$ $= 12 + 4 	imes 35.5 = 154 \ g/mol$.Moles of $CCl_4$ $(N)$ $= W_0 / M_0 = 158 / 154 \approx 1.026 \ mol$.Mass of solute $(W)$ $= 0.5 \ g$,Molar mass of solute $(M)$ $= 65 \ g/mol$.Moles of solute $(n)$ $= W / M = 0.5 / 65 \approx 0.00769 \ mol$.Using Raoult's Law for dilute solutions: $(p^0 - p) / p^0 = n / N$.$(143 - p) / 143 = 0.00769 / 1.026$.$(143 - p) / 143 \approx 0.007495$.$143 - p = 143 	imes 0.007495 \approx 1.0718$.$p = 143 - 1.0718 = 141.9282 \approx 141.93 \ mm \ Hg$.

At $25^\circ C$,the vapor 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 $CCl_4$,the vapor pressure of the resulting solution is $...... \ mm \ Hg$. (Density of $CCl_4 = 1.58 \ g/cm^3$)

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