A) 11.2mL
B) 22.4 mL
C) 33.6 mL
D) 44.8 mL
Correct Answer: D
Solution :
\[PbS+4{{H}_{2}}{{O}_{2}}\xrightarrow{{}}PbS{{O}_{4}}+4{{H}_{2}}O\] From the above equation \[\because \]1 mole of PbS required 4 moles of \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] \[\therefore \] 0.01 moles of PbS required 0.04 mole of \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] Weight of 0.04 mole \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}=1.36\,g\] 10 volume \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]means, 1 mL of such solution of \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]on decomposition by heat produces 10 mL of oxygen at NTP. \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]decomposes as, \[2{{H}_{2}}{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O+{{O}_{2}}\] Thus, 1 mL of 10 volume \[{{H}_{2}}{{O}_{2}}\]solution contains \[=\frac{68}{2400}\times 10g\]of \[{{H}_{2}}{{O}_{2}}\] \[=0.03035g\,{{H}_{2}}{{O}_{2}}\] \[\because \] 0.03035 \[\text{g}\,{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]is present in 1 mL of 10 volume \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] \[\therefore \] \[\,1.36\,\,g\,\,{{H}_{2}}{{O}_{2}}\]present in \[\frac{1}{0.03035}\times 1.36\,mL\] Of 10 volume \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] \[=44.81\,mL\]
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