Question

What would be the molality of $20\%(\mathrm{mass}/\mathrm{mass})$ aqueous solution of $\mathrm{NaOH}$?

Answer 1

Step 1: Determine the molar mass of $\mathrm{NaOH}$

$\mathrm{NaOH}=23+16+1=40\mathrm{g}/\mathrm{mol}$

Step 2: Determine the mass of $\mathrm{NaOH}$ and solvent

$20\%(\mathrm{mass}/\mathrm{mass})$ aqueous solution of $\mathrm{NaOH}$ means $20\mathrm{g}\mathrm{NaOH}$ in $100\mathrm{g}$ solution.

Therefore, mass of $\mathrm{NaOH}=20\mathrm{g}$

Mass of solvent $=80\mathrm{g}$

Step 3: Apply the formula of molality and calculate it

Molality $=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}}$

Molality $=\frac{20/40}{80/1000}$

Molality $=6.25\mathrm{mol}/\mathrm{kg}$

Hence, the molality of $20\%(\mathrm{mass}/\mathrm{mass})$ aqueous solution of $\mathrm{NaOH}$ is $6.25\mathrm{mol}/\mathrm{kg}$.