Question

The concentration of a solution of $H_2{O}_2$ is 6.8%(w/V) then the volume concentration of the solution is:-

• A. 22.4
• B. 11.2
• C. 20
• D. 5

Answer 1

The correct option is A 22.4

Given, $6.8g$ of $H_2{O}_2$ is present in $100mL$ of the solution $\left(6.8\%w/V\right).$

No. of moles of $H_2{O}_2$ is $n=\frac{mass}{Mol.mass\left(H_2{O}_2\right)}$

$\Rightarrow n=\frac{6.8}{34}=0.2mol$

And Volume of the $so{l}^n$, $V=100mL$ or $0.1L$

Now, Molarity of the solution,

$M=\frac{\text{mole of solute}}{\text{volume of solution in}\left(L\right)}$ = $\frac{0.2}{0.1}=2M$

We know that, Molarity $=\frac{Volume-Strength}{11.2}$

⇒ Volume strength of the solution $=Molarity\times 11.2$

⇒ Volume strength of the solution $=2\times 11.2=22.4$

Hence, option A is correct.