Question

What is molality of a $40\%\left(\text{w/w}\right)$ $H_2{O}_2$ aqueous solution?

• A. $19.61\text{m}$
• B. $18.52\text{m}$
• C. $15.50\text{m}$
• D. $12.82\text{m}$

Answer 1

The correct option is A $19.61\text{m}$

We know, $40\%\text{(w/w)}$ means $40\text{g}$ of $H_2{O}_2$ is present in $100\text{g}$ of the solution.
$\text{Therefore mass of solvent}=100\text{g}-40\text{g}=60\text{g}$
$\text{Number of moles of}{H}_2{O}_2=\frac{40\text{g}}{\text{molar mass of}{H}_2{O}_2}=\frac{40\text{g}}{34\text{g/mol}}=\frac{40}{34}\text{mol}$
$\text{Molality of solution}=\frac{\text{number of moles of}{H}_2{O}_2}{\text{mass of solvent in kg}}$
$=\frac{40}{60\times 34}\times 1000=19.61\text{m}$