A) 0.162
B) 0.675
C) 0.325
D) 0.486
Correct Answer: C
Solution :
[c] Molar mass of \[CHC{{l}_{3}}=119.5g/mole.\] Molar mass of \[C{{H}_{2}}C{{l}_{2}}=85g/mole.\] Moles of \[CHC{{l}_{3}}=\frac{11.95}{119.5}=0.1\,\,mole.\] Moles of \[C{{H}_{2}}C{{l}_{2}}=\frac{8.5}{85}=0.1\,\,mole.\] Mole fraction of \[CHC{{l}_{3}}=\frac{0.1}{0.2}=0.5\,\,mole.\] Mole fraction of \[C{{H}_{2}}C{{l}_{2}}=\frac{0.1}{0.2}=0.5\,\,mole.\] (Given - Vapour pressure of \[CHC{{l}_{3}}=200\,mm\] Hg = 0.263 atm. Vapour pressure of \[C{{H}_{2}}C{{l}_{2}}=415\,mm\] Hg = 0.546 atm.) (1 atm=760 mm Hg) \[\therefore \] \[{{P}_{\text{(above}\,\,\text{solution)}}}\] = Mole fraction of \[CHC{{l}_{3}}\times \] (Vapour pressure of \[CHC{{l}_{3}}\]) + Mole fraction of \[C{{H}_{2}}C{{l}_{2}}\times \] (Vapour pressure of \[C{{H}_{2}}C{{l}_{2}}\]) \[=0.5\times 0.263+0.5\times 0.546=0.4045\] Mole fraction of \[CHC{{l}_{3}}\] in vapour form \[=\frac{0.1315}{0.4045}=0.325.\]You need to login to perform this action.
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