A) 150,450 mm Hg
B) 125,150 mm Hg
C) 450,150 mm Hg
D) 250,300 mm Hg
Correct Answer: C
Solution :
Let vapour pressure of \[A=P_{A}^{0}\] Vapour pressure of \[B=P_{B}^{0}\] In first solution, Mole fraction of \[A({{x}_{A}})=\frac{1}{1+2}=\frac{1}{3}\] Mole fraction of\[B({{x}_{B}})=\frac{2}{1+2}=\frac{2}{3}\] According to Raoult's law, Total vapour pressure \[=250=P_{A}^{0}{{x}_{A}}+P_{B}^{0}{{x}_{B}}\] \[=250=\frac{1}{3}P_{A}^{0}+\frac{2}{3}P_{B}^{0}\] ? (i) In second solution Mole fraction of\[A({{x}_{A}})=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\] Mole fraction of \[B({{x}_{A}})=\frac{2}{4}=\frac{1}{2}\] \[\therefore \]Total vapour pressure \[=300=P_{A}^{0}{{x}_{A}}+P_{B}^{0}{{x}_{B}}\] \[300=\frac{1}{2}P_{A}^{0}+\frac{1}{2}P_{B}^{0}\]? (ii) Multiplying equation (i) by\[\frac{1}{2}\] and equation (ii) by\[\frac{1}{3}\] \[\frac{1}{6}P_{A}^{0}+\frac{2}{6}P_{B}^{0}=125\] \[\frac{\frac{1}{6}P_{A}^{0}+\frac{1}{6}P_{B}^{0}=100}{\frac{1}{6}P_{B}^{0}=25}\] \[P_{B}^{0}=25\times 6=150mm\,Hg\] On substituting value of \[P_{B}^{0}\]in equation (ii) we get\[\lambda =\frac{h}{mv}\] ?(i) \[300=P_{A}^{0}\times \frac{1}{2}+150\times \frac{1}{2}\] \[P_{A}^{0}=450mmHg\]
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