Question

Carbon tetrachloride $\left(CC{l}_4\right)$ is mixed in 390 g of benzene to make a solution for supporting an organic chemical reaction in which mole fraction of benzene is 0.91. Find the molality of the $CC{l}_4$ solution.

• A. $2.67\text{m}$
• B. $4.25\text{m}$
• C. $1.26\text{m}$
• D. $4.44\text{m}$

Answer 1

The correct option is C $1.26\text{m}$

Given:
Mass of benzene (solvent) = 390 g
Mole fraction of the solute + mole fraction of the solvent = 1
$Molefractionofsolute=x_{CC{l}_4}=1-0.91=0.09$
$x_{CC{l}_4}=\frac{molesofsolute\left(n_{CC{l}_4}\right)}{molesofsolute\left(n_{CC{l}_4}\right)+molesofsolvent\left(n_{benzene}\right)}$
$x_{\left(CC{l}_4\right)}=0.09=\frac{n_{CC{l}_4}}{(n_{CC{l}_4}+n_{benzene})}=\frac{n_{\left(CC{l}_4\right)}}{\left(n_{CC{l}_4}\right)+390/78}$
On solving, $n_{CC{l}_4}=0.495$
$Molality\left(m\right)=\frac{molesofsolute}{massofsolventing}\times 1000$
$Molality=\frac{0.495}{390}\times 1000=1.26m$