An ideal solution of liquid ApA∗=100mmHg and liquid BpB∗=30mmHg has 0.6 mole fraction of A in the vapour phase. The mole fraction of A in the liquid phase will be

# An ideal solution of liquid $\mathrm{A}\left({p}_{\mathrm{A}}^{\ast }=100\mathrm{mmHg}\right)$ and liquid $\mathrm{B}\left({p}_{\mathrm{B}}^{\ast }=30\mathrm{mmHg}\right)$ has 0.6 mole fraction of A in the vapour phase. The mole fraction of A in the liquid phase will be

1. A

0.21

2. B

0.25

3. C

0.31

4. D

0.41

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### Solution:

We have  ${y}_{\mathrm{A}}=\frac{{p}_{\mathrm{A}}}{{p}_{\mathrm{A}}+{p}_{\mathrm{B}}}=\frac{{x}_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }}{{x}_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }+\left(1-{x}_{\mathrm{A}}\right){p}_{\mathrm{B}}^{\ast }}=\frac{{x}_{\mathrm{A}}{p}_{\mathrm{A}}^{\ast }}{{p}_{\mathrm{B}}^{\ast }+\left({p}_{\mathrm{A}}^{\ast }-{p}_{\mathrm{B}}^{\ast }\right){x}_{\mathrm{A}}}$

This gives  ${x}_{\mathrm{A}}=\frac{{y}_{\mathrm{A}}{p}_{\mathrm{B}}^{\ast }}{{p}_{\mathrm{A}}^{\ast }+\left({p}_{\mathrm{B}}^{\ast }-{p}_{\mathrm{A}}^{\ast }\right){y}_{\mathrm{A}}}=\frac{\left(0.69\right)\left(30\right)}{100+\left(30-100\right)\left(0.60\right)}=0.31$

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