A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{9}\]
D) \[\frac{8}{9}\]
Correct Answer: C
Solution :
Let w kg of \[C{{H}_{4}}\] and \[H_{2}^{.}\] are taken in a container. \[\therefore \] \[{{n}_{C{{H}_{4}}}}=\frac{w}{16}\] and \[{{n}_{C{{H}_{2}}}}=\frac{w}{2}\] and mole fraction of \[C{{H}_{4}}({{X}_{C{{H}_{4}}}})=\frac{\frac{w}{1.6}}{\frac{w}{16}+\frac{w}{2}}=\frac{1}{9}\] \[\therefore \] \[{{p}_{C{{H}_{4}}}}={{p}_{Total}}\times {{X}_{C{{H}_{4}}}}\] \[={{p}_{Total}}\times \frac{1}{9}\]
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