Question

Calculate the molality of $20\%\left(w/w\right)H_2{O}_2\left(aq\right)$ solution?

• A. $12.66m$
• B. $1.56m$
• C. $7.35m$
• D. $19.31m$

Answer 1

The correct option is C $7.35m$

$20\%\left(w/w\right)$ means $20g$ of $H_2{O}_2$ is present in $100g$ of the solution.
$\text{Mass of solvent(water)}=100g-20g=80g$

$\text{Number of moles of}{H}_2{O}_2=\frac{20g}{\text{molar mass of}{H}_2{O}_2}=\frac{20g}{34g/mol}=\frac{20}{34}mol$
$\text{Molality of solution}=\frac{\text{Number of moles of}{H}_2{O}_2}{\text{Mass of solvent in kg}}$
$=\frac{20}{80\times 34}\times 1000=7.35m$