• # question_answer Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapour pressures of pure A and pure B are $7\times {{10}^{3}}$ and $12\times {{10}^{3}}$ Pa, respectively. The composition of the vapour in equilibrium with a solution containing 40 mole per cent of A at this temperature is: A) ${{x}_{A}}=0.37;{{x}_{B}}=0.63$     B) ${{x}_{A}}=0.28;\,{{x}_{B}}=0.72$ C) ${{x}_{A}}=0.4;\,\,{{x}_{B}}=0.6$ D) ${{x}_{A}}=0.76;6{{x}_{B}}=0.24$

 ${{P}_{A}}={{\chi }_{A}}P_{A}^{0}={{Y}_{A}}{{P}_{T}}$ ${{P}_{T}}={{\chi }_{A}}P_{A}^{0}+{{\chi }_{B}}P_{B}^{0}$ $=0.4\times 7\times {{10}^{3}}+0.6\times 12\times {{10}^{3}}={{10}^{4}}$ $0.4\times 7\times {{10}^{3}}={{Y}_{A}}\times {{10}^{4}}$ ${{Y}_{A}}=0.28$ ${{Y}_{B}}=1-0.28=0.72.$