Question

Calculate the molality of $25.5\%$ (w/w) $H_2{O}_2$ aqueous solution? (Give your answer upto two decimal places.)

• A. $10.06m$
• B. $1.06m$
• C. $100.06m$
• D. $106.01$

Answer 1

The correct option is A $10.06m$

$25.5\%$ (w/w) means $25.5\text{g}$ of $H_2{O}_2$ is present in $100\text{g}$ of the solution.
Mass of solvent $=100\text{g}-25.5\text{g}=74.5\text{g}$
Number of moles of $H_2{O}_2=\frac{25.5g}{\text{molar mass of}{H}_2{O}_2}=\frac{25.5g}{34g/mol}=\frac{25.5}{34}mol$
$\text{Molality of solution}=\frac{\text{number of moles of}{H}_2{O}_2}{\text{mass of solvent in kg}}$
$=\frac{25.5}{74.5\times 34}\times 1000=10.06m$