AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

For vessel A, The reaction of hydrogen peroxide is shown below, $2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O} + \mathrm{O}_2$ The $10 \text{ mL} \mathrm{O}_2$ is obtained by, $w = \frac{2 \text{ mol of } \mathrm{H}_2\mathrm{O}_2 \times V_{\mathrm{H}_2\mathrm{O}_2}}{V_{\text{at NTP}}}$ Substitute the value in above we get, $w = \frac{2 \times 34 \times 10}{22400} = 0.03 \text{ g} \mathrm{H}_2\mathrm{O}_2$ For vessel B, The solution of 30 mL $\mathrm{O}_2$ is obtained by, $w = \frac{2 \text{ mol of } \mathrm{H}_2\mathrm{O}_2 \times V_{\mathrm{H}_2\mathrm{O}_2}}{V_{\text{at NTP}}}$ Substitute the value in above we get, $w = \frac{2 \times 34 \times 30}{22400} = 0.09 \text{ g} \mathrm{H}_2\mathrm{O}_2$ For vessel B, The molarity is give as, $M = \frac{n_{\mathrm{H}_2\mathrm{O}_2}}{V_{\text{sol}}}$ Here, the number of moles of $\mathrm{H}_2\mathrm{O}_2$ is $n_{\mathrm{H}_2\mathrm{O}_2}$ and the volumeoff the solution is $V_{\text{sol}}$ Substitute the value in above expression. $2 = \frac{\frac{x}{34}}{\frac{250}{1000}}$ $x = 17 \text{ g}$ Thus, the order of weight of $\mathrm{H}_2\mathrm{O}_2$ in vessel is $C > B > A$.