Question

A solution is prepared by dissolving 31.9 g cesium chloride in 60.0 g water. The volume of the solution is 63.3 mL. Calculate the molality of the CsCl solution.

Answer 1

Answer: The molality of the CsCl solution is 3.2 m.

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molality=\frac{n\times 1000}{W_s}$

where,

n = moles of solute

moles of solute =$\frac{\text {given mass}}{\text {molar mass}}=\frac{31.9g}{168.36g/mol}=0.19moles$

Now put all the given values in the formula of molarity, we get

$W_s$ = weight of solvent (water) in g  = 60.0 g

Molality =$\frac{0.19\times 1000}{60}=3.2mol/kg$

Thus molality of the CsCl solution is 3.2 m.