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

Step-by-step 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* 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)* 1000}{M_(solute)* 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=(24.8* 1000)/(40* 1125.2)\\\\\text{Molality of }NaOH=0.551m`

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