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  • Vapour pressure of mixture of liquid A and liquid B at is given by <img alt="\mathrm{P_T=360 x_B+8

Vapour pressure of mixture of liquid A and liquid B at 70^{\circ} \mathrm{C} is given by \mathrm{P_T=360 x_B+82 (in \; mm)} Where \mathrm{x_B} is the mole fraction of B, in the liquid mixture. Find vapor pressure pure A and Pure B

Option: 1

\begin{aligned} & \mathrm{\rho ^{\circ} A=82 \; \mathrm{mmHg}, \rho ^{\circ} B=442\; \mathrm{mmHg}} \\ \end{aligned}


Option: 2

\begin{aligned} & \mathrm{{P}^{\circ} _A=90 \mathrm{mmHg}, P^{\circ} _B=270 \mathrm{~mm} \mathrm{H} \mathrm{g}} \\ \end{aligned}


Option: 3

\begin{aligned} & \mathrm{{P}^{\circ} _A=50 \mathrm{mmHg}, P^{\circ} _B=360 \mathrm{~mm} \mathrm{H} \mathrm{g}} \\ \end{aligned}


Option: 4

\begin{aligned} & \mathrm{{P}^{\circ} _A=60 \mathrm{mmHg}, P^{\circ} _B=250 \mathrm{~mm} \mathrm{H} \mathrm{g}} \\ \end{aligned}


Answers (1)

best_answer

\mathrm{\rho_T=360 \: x_B+82}

When \mathrm{x_B=0, x_A=1}

\begin{aligned} & \mathrm{P_T=360 \times 0+82} \\ \\& \mathrm{P^{\circ} _A}=82 \mathrm{~mm} \mathrm{Hg} \end{aligned}

When \mathrm{x_B=1 \text { and } x_A=0}

\begin{aligned} & \mathrm{P_T=360 \times 1+82} \\ \\& \mathrm{P^{\circ} _B=442 \: \mathrm{mmHg}} \end{aligned}

Posted by

himanshu.meshram

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