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The vapour pressure of two pure liquids $A$ and $B$ that form an ideal solution, are 400 and $800 \mathrm{~mm}$ of Hg respectivelyat a temperature $t^{\circ} \mathrm{C}$. The mole fraction of $A$ in a solution of $A$ and $B$ whose boiling point is $t^{\circ} \mathrm{C}$ will be
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The correct answer is:
0.1
V.P. of solution at $t^{\circ} \mathrm{C}=760 \mathrm{~mm}$ [at b.p., V.P. of solution =atompheric pressure] Thus $=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{\infty} \mathrm{x}_{\mathrm{B}}$
or $\mathrm{P}=\mathrm{P}_{\mathrm{A}}^{0} \cdot \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0}\left(1-\mathrm{x}_{\mathrm{A}}\right)\left[\because \mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}=1\right]$
or $760=400 \mathrm{X}_{\mathrm{A}}+800\left(1-\mathrm{X}_{\mathrm{A}}\right)[\because \mathrm{P}=760 \mathrm{mmofHg}]$
or $-800+760=-400 x_{\mathrm{A}}$
or $-40=-400 \mathrm{x}_{\mathrm{A}}$
or $x_{\mathrm{A}}=\frac{40}{400}=0.1$
Thus mole fraction in solution is 0.1
or $\mathrm{P}=\mathrm{P}_{\mathrm{A}}^{0} \cdot \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0}\left(1-\mathrm{x}_{\mathrm{A}}\right)\left[\because \mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}=1\right]$
or $760=400 \mathrm{X}_{\mathrm{A}}+800\left(1-\mathrm{X}_{\mathrm{A}}\right)[\because \mathrm{P}=760 \mathrm{mmofHg}]$
or $-800+760=-400 x_{\mathrm{A}}$
or $-40=-400 \mathrm{x}_{\mathrm{A}}$
or $x_{\mathrm{A}}=\frac{40}{400}=0.1$
Thus mole fraction in solution is 0.1
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