Q21 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Chemistry

At \(T(\mathrm{~K})\), the vapour pressures of pure liquids \(A\) and \(B\) are \(100 \mathrm{~mm}\) and \(160 \mathrm{~mm}\) respectively. An ideal solution is formed by mixing 2 moles of \(A\) and 3 moles of \(B\) at the same temperature. The mole fraction of \(A\) and \(B\) in the vapour state respectively are

  • A
    \(0.706,0.294\)
  • Correct answer
    B
    \(0.294,0.706\)
  • C
    \(0.40,0.60\)
  • D
    \(0.60,0.40\)

Solution

Key Idea Vapour pressure of solution, \(p_{\text {total }}=p_A+p_{B^{\prime}}=\chi_A p_A^{\circ}+\chi_B p_B^{\circ} \quad\left[\because p_A=\chi_A p_A^{\circ}\right]\) Also vapour pressure of component \(1, p_1=y_1 p_{\text {total }}\) where \(y\), is the mole fraction of component 1 in vapour phase. Given, Vapour pressure of pure liquid \(A, p_A^{\circ}=100 \mathrm{~mm}\) Vapour pressure of pure liquid \(B, p_B^{\circ}=160 \mathrm{~mm}\) \(\therefore\) Total vapour pressure of solution \(=p_A+p_B\) \(\begin{aligned} p_{\text {total }} & =\chi_A p_A^{\circ}+\chi_B p_B^{\circ}=\frac{2}{5} \times 100+\frac{3}{5} \times 160 \\ & =40+96=136 \mathrm{~mm} \end{aligned}\) Also \(\quad p_A=y_A p_{\text {total }}\) where, \(y_A\) is the mole fraction of \(A\). Mole fraction of \(A, y_A=\frac{p_A}{p_{\text {total }}}=\frac{40}{136}=0.294\) \(\therefore \quad y_B=1-y_A=1-0.294=0.706\)

Hence, the correct answer is option: 'B'.

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