Q. In a solution of 7.8 g benzene $(C_{6}H_{6})$ and 46.0 g toluene $(C_{6}H_{5}CH_{3})$, the mole-fraction of benzene is
A) $\frac{1}{2}$
B) $\frac{1}{3}$
C) $\frac{1}{5}$
D) $\frac{1}{6}$
Explanation
(d)
Mole fraction of $C_{6}H_{6} = \frac{\frac{7.8}{78}}{\frac{7.8}{78} + \frac{46}{92}} = \frac{1}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.