Question

What is final molarity of $30\%$ (w/w) $H_2{O}_2$ aqueous solution, having density $1.11\text{g/ml}$, if the solution is diluted to triple the initial volume?

• A. $9.8\text{M}$
• B. $4.85\text{M}$
• C. $12.6\text{M}$
• D. $3.26\text{M}$

Answer 1

The correct option is D $3.26\text{M}$

$30\%$ (w/w) means $30\text{g}$ of $H_2{O}_2$ is present in $100\text{g}$ of the solution.
Mass of solvent = $100\text{g}-30\text{g}=70\text{g}$
Number of moles of $H_2{O}_2=\frac{30\text{g}}{\text{molar mass of}{H}_2{O}_2}=\frac{30g}{34g/mol}=\frac{30}{34}mol$
$\text{Volume of initial solution}\left(V_1\right)=\frac{\text{mass of solution}}{\text{density of solution}}=\frac{100g}{1.11g/ml}\simeq 90\text{ml}=0.09\text{L}$
$\text{Molarity of initial solution}\left(M_1\right)=\frac{\text{number of moles of}{H}_2{O}_2}{\text{volume of solution in L}}=\frac{30}{0.09\times 34}=9.8M$
Now, since the number of moles of $H_2{O}_2$ remain same, before and after the dilution.
$\therefore M_1{V}_1=M_2{V}_2$, where $M_2$ and $V_2$ are final molarity and volume, after dilution, respectively.
$\therefore M_2=\frac{9.8\times 90}{3\times 90}\simeq 3.26\text{M}$