A) 0.99
B) 0.11
C) 0.19
D) 0.81
Correct Answer: B
Solution :
Let us consider total mass of mixture is 100 g in which 30 g is \[CC{{l}_{4}}\] and 70 g is ethanol. \[\operatorname{Moles} of CC{{l}_{4}}\,\,=\,\,\frac{30}{154} =\,\,0.1948\] \[\operatorname{Moles} of ethanol =\,\,\frac{70}{46} = 1.52\] \[\operatorname{Mole} fraction of CC{{l}_{4}} =\,\,\frac{0.1948}{0.1948+1.52}\]\[=\,\,\,\,0.1136\]You need to login to perform this action.
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