Question

The strengths of $5.6$ volume hydrogen peroxide (of density $1\mathrm{g}/\mathrm{mL}$) in terms of mass percentage and molarity $\left(\mathrm{M}\right)$, respectively, are: (Take the molar mass of hydrogen peroxide as $34\mathrm{g}/\mathrm{mol}$)

• A. $0.85\mathrm{and}0.5$
• B. $0.85\mathrm{and}0.25$
• C. $1.7\mathrm{and}0.25$
• D. $1.7\mathrm{and}0.5$

Answer 1

The correct option is D

$1.7\mathrm{and}0.5$

The explanation for the correct option:

D: $1.7\mathrm{and}0.5$

The expression used for the calculation of molarity:

$\begin{array}{rcl}\mathrm{Molarity},\mathrm{M}& =& \frac{\mathrm{n}}{\mathrm{v}}\\ \mathrm{where},\mathrm{M}& =& \mathrm{molar}\mathrm{concentration}\\ \mathrm{n}& =& \mathrm{moles}\mathrm{of}\mathrm{solute}\\ \mathrm{v}& =& \mathrm{liters}\mathrm{of}\mathrm{solution}\\ \mathrm{Volume}\mathrm{strength}& =& 5.6\mathrm{V}\\ \mathrm{Molarity}& =& 5.6/11.2\\ & =& 0.5\mathrm{mol}/\mathrm{l}\end{array}$

Calculation of mass percentage,

$\begin{array}{rcl}& & \mathrm{Assuming}1\mathrm{litre}\mathrm{solution}\\ \mathrm{mass}\mathrm{of}\mathrm{solution}& =& 1000\mathrm{ml}\times 1\mathrm{g}/\mathrm{ml}=1000\mathrm{g}\\ \mathrm{mass}\mathrm{of}\mathrm{solute}& =& \mathrm{moles}\times \mathrm{molar}\mathrm{mass}\\ & =& 0.5\mathrm{mol}\times 34\mathrm{g}/\mathrm{mol}\\ & =& 17\mathrm{gm}.\\ & \Rightarrow & \mathrm{mass}\%=\frac{17}{100}\times 100\\ & =& 1.7\%\end{array}$

Therefore, Option D $1.7\mathrm{and}0.5$ is the correct option.