Question

What is molality of 30% (w/w) $H_2{O}_2$ aqueous solution?

• A. 8.82 m
• B. 12.60 m
• C. 15.60 m
• D. 10.67 m

Answer 1

The correct option is B 12.60 m

30% (w/w) means 30 g of $H_2{O}_2$ is present in 100 g of the solution.
Mass of solvent = 100 g - 30 g = 70 g
Number of moles of $H_2{O}_2=\frac{30g}{molarmassof{H}_2{O}_2}=\frac{30g}{34g/mol}=\frac{30}{34}mol$
$Molalityofsolution=\frac{numberofmolesof{H}_2{O}_2}{massofsolventinkg}$
$=\frac{30}{70\times 34}\times 1000=12.60m$