Solutions Question 392
Question: When 2g of non-volatile hydrocarbon containing 94.4 percent carbon is dissolved in 100g benzene, the vapour pressure of Benzene is lowered from 74.66 torr to 74.01 torr. Determine the molecular formula of the hydrocarbon.
Options:
A) $ C_6H_6 $
B) $ C_{12}H_6 $
C) $ C_7H_8 $
D) $ C_{14}H_{10} $
Show Answer
Answer:
Correct Answer: D
Solution:
According to Raoult’s law $ \frac{\Delta p}{p^{0}}=x_2 $ Where $ -\Delta p $ = (74.01 - 74.66) torr and $ p^{0} $ = 74.66 torr
If M is the molar mass of hydrocarbon, then $ X_2=\frac{n_2}{n_1+n_2}=\frac{\frac{8}{M}}{( \frac{100}{78} )+( \frac{2}{M} )} $
Hence $ \frac{74.66-74.01}{74.66}=\frac{\frac{2}{M}}{\frac{100}{78}+\frac{2}{M}} $
Solving for M, we get, M = 177.6 g $ mo{l^{-1}} $ . Given mass ratio is $ m_{C}:m_{H} $ : : 94.4 : 5.6 This atomic ratio is $ N_{C}:N_{H}::\frac{84.4}{12} :\frac{5.6}{1}\Rightarrow 7.87 :5.6 $
$ \Rightarrow 1.4 :1\Rightarrow 7:5 $
Hence, Empirical formula is $ C_7H_5 $
Molar Empirical mass = 89 g $ mo{l^{-1}} $
Number of $ C_7H_5 $ unit in the given molecule $ =\frac{Molar,mass}{Molar,empirical,mass}=\frac{177.6}{89}\cong 2 $
Thus molecular formula is $ C_{14}H_{10} $
