Question

A sodium hydroxide solution that contains 24.8 grams of NaOH per L of solution has a density of 1.15 g/mL. Calculate the molality of the NaOH in this solution.

Answer 1

Answer: The molality of NaOH in the solution is 0.551 m

Explanation:

To calculate mass of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.15 g/mL

Volume of solution = 1 L = 1000 mL     (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

$1.15 g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.15g/mL\times 1000mL)=1150g$

We are given:

Mass of solute (NaOH) = 24.8 grams

Mass of solution = 1150 grams

Mass of solvent = Mass of solution - mass of solute = [1150 - 24.8] g = 1125.2 g

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (NaOH) = 24.8 g

$M_{solute}$ = Molar mass of solute (NaOH) = 40 g/mol

$W_{solvent}$ = Mass of solvent = 1125.2 g

Putting values in above equation, we get:

$\text{Molality of }NaOH=\frac{24.8\times 1000}{40\times 1125.2}\\\\\text{Molality of }NaOH=0.551m$

Hence, the molality of NaOH in the solution is 0.551 m