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

Step-by-step 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)* 1000}{M_(solute)* 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}=(3.301* 1000)/(56.1* 96.699)\\\\\text{Molality of KOH}=0.608m`

Hence, the molality of potassium hydroxide solution is 0.608 m