Question

An aqueous sodium acetate, NaC2H3O2 , solution is made by dissolving 0.395 mol NaC2H3O2 in 0.505 kg of water. Calculate the molality of the solution.

Answer 1

Answer: The molality of $NaC_2H_3O_2$ solution is 0.782 m

Explanation:

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molality:

$\text{Molality of solution}=\frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}}$ .....(1)

Given values:

Moles of $NaC_2H_3O_2$ = 0.395 mol

Mass of solvent (water) = 0.505 kg

Putting values in equation 1, we get:

$\text{Molality of }NaC_2H_3O_2=\frac{0.395mol}{0.505kg}\\\\\text{Molality of }NaC_2H_3O_2=0.782m$

Hence, the molality of $NaC_2H_3O_2$ solution is 0.782 m