Q9 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Chemistry

What is the \% strength of 22.4 volume of $\mathrm{H}_2 \mathrm{O}_2$ solution?

  • A
    $3.4 \%$
  • B
    $2.5 \%$
  • C
    $5 \%$
  • Correct answer
    D
    $6.8 \%$

Solution

$22.4 \mathrm{vol} \mathrm{H} \mathrm{O}_2$ means $\mathrm{l} \mathrm{mL}$ of $\mathrm{H}_2 \mathrm{O}_2$ solution gives $22.4 \mathrm{~mL} \mathrm{O}_2$. $\begin{array}{ll} 2 \mathrm{H}_2 \mathrm{O}_2 & \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\ 2 \mathrm{~mol} & 1 \mathrm{~mol} \\ 68 \mathrm{~g} & 22400 \mathrm{~mL} \text { at NTP } \end{array}$ $\therefore 22400 \mathrm{~mL} \mathrm{O}_2$ is obtained by $68 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2$ $\therefore 22.4 \mathrm{~mL} \mathrm{O}_2$ is obtained by $\frac{68 \times 22.4}{22400}=0.068 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2$ or $1 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}_2$ solution $=0.068 \mathrm{~g}$ $\therefore 100 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}_2$ solution $=0.0680 \times 100=6.8 \%$

Hence, the correct answer is option: 'D'.

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