Question

Carbon tetrachloride 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 m
• B. 4.25 m
• C. 1.26 m
• D. 4.44 m

Answer 1

The correct option is C 1.26 m

In the given solution, $CC{l}_4$ is the solute and benzene is the solvent.
Given,
Mass of benzene = 390 g
$\text{No. of moles of benzene}=\frac{Mass}{Molecularmass}=\frac{390}{78}=5mol$

$\text{Mole fraction of benzene,}{X}_{benzene}=0.91$
$\text{Mole fraction of solvent}=\frac{No.ofmolesofsolvent}{No.ofmolesofsolute+No.ofmolesofsolvent}\text{Mole fraction of benzene}=\frac{No.ofmolesofbenzene}{No.ofmolesofCC{l}_4+No.ofmolesofbenzene}0.91=\frac{5}{n_{CC{l}_4}+5}0.91n_{CC{l}_4}+4.55=5n_{CC{l}_4}=\frac{0.45}{0.91}{n}_{CC{l}_4}=0.494mol$
$\text{Molality of solution}=\frac{No.ofmolesofsolute}{Massofsolventinkg}\text{Molality of solution}=\frac{0.494}{390}\times 1000$
Molality = 1.26 m