Question

A solution is prepared by dissolving 52.3 g of caesium chloride in 60.0 g water. The volume of the solution is 63.3 ml. Calculate the mass percent, molarity, molality, and mole fraction of the $CsCl$ solution.

Short answer

Molarity is$4.91mol/L$.

Molality is 5.81 mol/kg.

Mass percent is 46.6%

Mole percent is 0.0854

Step-by-step solution

Step 1:

Find $n\left(H_{2}O\right)$.

$n\left(H_{2}O\right)=\frac{60.0g}{18.016g/mol}=3.33mol$.

Now find $n\left(CsCl\right)$.

$n\left(CsCl\right)=\frac{52.3g}{168.36g/mol}$.

Calculate molarity.

$M=\frac{0.311mol}{0.0633L}=4.91mol/L$

Step 2:

Calculate molality.

$m=\frac{0.311mol}{0.0600kg}=5.18mol/kg$.

Step 3:

Calculate mole percent.

$\chi \left(CsCl\right)=\frac{0.311mol}{0.311mol+3.33mol}=0.0854$.

Step 4:

Calculate mass percent.

$=\frac{52.3g}{52.3g+60.0g}\times 100=46.6\%$