Question

A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give your answer to 3 decimal places.

Answer 1

Answer: The molality of potassium hydroxide solution is 0.608 m

Explanation:

We are given:

3.301 mass % of potassium hydroxide solution.

This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution

Mass of solvent = Mass of solution - Mass of solute (KOH)

Mass of solvent = (100 - 3.301) g = 96.699 g

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

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

Where,

$m_{solute}$ = Given mass of solute (KOH) = 3.301 g

$M_{solute}$ = Molar mass of solute (KOH) = 56.1 g/mol

$W_{solvent}$ = Mass of solvent = 96.699 g

Putting values in above equation, we get:

$\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m$

Hence, the molality of potassium hydroxide solution is 0.608 m