Question

Commercial $11.2$ volume $H_2{O}_2$ solution has a molarity of

• A. $1.0$
• B. $0.5$
• C. $11.2$
• D. $1.12$

Answer 1

Answer: (A)

$10$Vol. $H_2{O}_2=3.035g{H}_2{O}_2$
$\therefore 11.2$ vol. $H_2{O}_2=\frac{3.035\times 11.2}{10}=33.99g$
molarity $=\frac{\text{Strength}}{\text{mol.wt}}=\frac{33.09}{34}=1M$