At 298 K,0.714 moles of liquid A is dissolve | vedclass

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(D) According to Raoult's law for a solution of two volatile liquids,the total vapour pressure $P_{total}$ is given by: $P_{total} = P_A^0 \chi_A + P_

Vedclass · Accepted Answer

$500$

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(D) According to Raoult's law for a solution of two volatile liquids,the total vapour pressure $P_{total}$ is given by: $P_{total} = P_A^0 \chi_A + P_B^0 \chi_B$. Given: $n_A = 0.714 \ mol$,$n_B = 5.555 \ mol$. Total moles $n_{total} = 0.714 + 5.555 = 6.269 \ mol$. Mole fractions are: $\chi_A = \frac{0.714}{6.269} \approx 0.1139$ and $\chi_B = \frac{5.555}{6.269} \approx 0.8861$. Given $P_{total} = 475 \ torr$ and $P_A^0 = 280.7 \ torr$. Substituting values: $475 = (280.7 	imes 0.1139) + (P_B^0 	imes 0.8861)$. $475 = 31.97 + (P_B^0 	imes 0.8861)$. $443.03 = P_B^0 	imes 0.8861$. $P_B^0 = \frac{443.03}{0.8861} \approx 499.98 \ torr \approx 500 \ torr$.

At $298 \ K$,$0.714$ moles of liquid $A$ is dissolved in $5.555$ moles of liquid $B$. The vapour pressure of the resultant solution is $475 \ torr$. The vapour pressure of pure liquid $A$ at the same temperature is $280.7 \ torr$. What is the vapour pressure of pure liquid $B$ in $torr$?

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