Question

A sample of $H_2{O}_2$ solution labelled as "$28$ volume" has density of $265\text{g/L}$. Mark the correct option(s) representing concentration of same solution in other units:

• A. $\text{Molarity of}{H}_2{O}_2\left(M_{H_2{O}_2}\right)=2.5$
• B. $\text{\%}\frac{w}{v}=17$
• C. $\text{Mole fraction of}{H}_2{O}_2=0.2$
• D. $\text{Molality of}{H}_2{O}_2\left(m_{H_2{O}_2}\right)=13.88$

Answer 1

Answer: (A), (C), (D)

$\text{Molality of}{H}_2{O}_2\left(m_{H_2{O}_2}\right)=13.88$

Given,
$\therefore \text{Molarity}=\frac{\text{volume strength}}{11.2}=\frac{28}{11.2}=2.5\text{M}$
$\therefore 1\text{L}$ contain $2.5$ moles of $H_2{O}_2$
$\text{Mass of}{H}_2{O}_2\text{in 1 L solution}\left(w_{H_2{O}_2}\right)=2.5\times 34\text{g}=85\text{g}{H}_2{O}_2$
$\text{Given, density of the solution}\left(d\right)=265\text{g/L}$
$\text{Therefore weight of}1\text{litre solution}=265\text{g}$
$\therefore w_{H_{2}O}=265-85\text{g}=180\text{g}{H}_{2}O=10\text{mol}$
$x_{H_2{O}_2}=\frac{2.5}{2.5+10}=0.2$
$\text{\%}\frac{w}{v}=\frac{2.5\times 34}{1000}\times 100=8.5$
$m_{H_2{O}_2}=\frac{2.5}{180}\times 1000=13.88\text{m}$