Question

Carbon tetrachloride $\left(CC{l}_4\right)$ is mixed in $390\text{g}$ 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) $=390\text{g}$
Mole fraction of the solute + mole fraction of the solvent $=1$
$\text{Mole fraction of solute}=x_{CC{l}_4}=1-0.91=0.09$
$x_{CC{l}_4}=\frac{\text{moles of solute}\left(n_{CC{l}_4}\right)}{\text{moles of solute}\left(n_{CC{l}_4}\right)+\text{moles of solvent}\left(n_{\text{benzene}}\right)}$
$x_{\left(CC{l}_4\right)}=0.09=\frac{n_{CC{l}_4}}{(n_{CC{l}_4}+n_{\text{benzene}})}=\frac{n_{\left(CC{l}_4\right)}}{\left(n_{CC{l}_4}\right)+390/78}$
On solving, $n_{CC{l}_4}=0.495$
$\text{Molality (m)}=\frac{\text{moles of solute}}{\text{mass of solvent in g}}\times 1000$
$\text{Molality}=\frac{0.495}{390}\times 1000=1.26\text{m}$