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The molarity of \mathrm{H}_2 \mathrm{O}_2 in a commercial sample of \mathrm{H}_2 \mathrm{O}_2 marked as 22.4 V and the mass of \mathrm{O}_2 available from 100 mL of the same sample are respectively

Option: 1

1.5 M, 2.4 g


Option: 2

2 M, 3.2 g


Option: 3

3 M, 4.8 g 


Option: 4

1 M, 1.6 g 


Answers (1)

best_answer

\mathrm{H}_2 \mathrm{O}_2 \text { (aq) } \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})+ \frac{1}{2} \mathrm{O}_2\mathrm{(g)}

\begin{aligned} 1 \text{M} & =0.5 \mathrm{~mol} \text { of } \mathrm{O}_2 \\ & =11.2 \mathrm{~L} \text { (STP) } \end{aligned}

Hence, 

\text { molarity }=\frac{22.4 \mathrm{~L}}{11.2 \mathrm{~L}}=2 \mathrm{M}

Number of moles of \mathrm{H}_2 \mathrm{O}_2  in 100 mL

\begin{aligned} & =\mathrm{M \times V} \\ & =2 \times\left(100 \times 10^{-3}\right) \\ & =0.2 \end{aligned}

Mass of \mathrm{O_{2}} obtained 

=0.2 \times \frac{32}{2}

=3.2\; \text{g}

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Rishi

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