Question

The cryoscopic constant for acetic acid is $3.6 K \,kg/mol$. A solution of $1\, g$ of a hydrocarbon in $100 \,g$ of acetic acid freezes at $16.14^{\circ}C$ instead of the usual $16.60^{\circ}C$. The hydrocarbon contains $92.3\%$ carbon. What is the molecular formula?

• A. $C_{6}H_{6}$
• B. $C_{6}H_{12}$
• C. $C_{8}H_{18}$
• D. $C_{10}H_{6}$

Answer 1

Answer: (A)

$K_{b}=3.6\,K \,kg /mol$
$\Delta T_{f} =16.6-16.14=0.46^{\circ}C$
$\Delta T_{f}=\frac{1000\,k_{f} w_{1}}{m_{1} w_{2}}$
$0.46=\frac{1000\times3.6\times1}{m_{1}\times100}$
$m_{1}=\frac{36}{0.46}=78.26\,g$
$C=\frac{92.3}{100}\times78.26=72.23\,g$
$H=\frac{7.7\times78.26}{100}=6.026\,g$
Whole ratio
$C=\frac{72.23}{12}=6.02$
$H=\frac{6.026}{1}=6.02$
Therefore, the molecular formula is $C_{6}H_{6}$