A) \[8.4\]
B) \[4.8\]
C) \[5.2\]
D) \[8.8\]
Correct Answer: A
Solution :
We know that, strength = normality \[\times \]equivalent weight of \[{{H}_{2}}{{O}_{2}}\] \[=1.5\times 17\] \[\therefore \]Strength of \[1.5N\,\,{{H}_{2}}{{O}_{2}}\] solution \[=25.5\text{ }g/L\]. Now, \[68\text{ }g\] of \[{{H}_{2}}{{O}_{2}}\]gives \[22400mL\] \[{{O}_{2}}\] at NTP \[\therefore \] \[25.5g\] of \[{{H}_{2}}{{O}_{2}}\] gives \[\frac{22400}{68}\times 25.5mL\] \[{{O}_{2}}\] at NTP \[=8400\text{ }mL\]of \[{{O}_{2}}\] at NTP But, \[25.5\text{ }g\]of \[{{H}_{2}}{{O}_{2}}\] are present in \[1000\text{ }mL\]of \[{{H}_{2}}{{O}_{2}}\]solution. Hence, \[1000\text{ }mL\]of \[{{H}_{2}}{{O}_{2}}\] solution gives \[8400\text{ }mL\]of \[{{O}_{2}}\] at NTP \[\therefore \] \[1\text{ }mL\]of \[{{H}_{2}}{{O}_{2}}\] solution will give \[=\frac{8400}{1000}\] \[=8.4mL\]of \[{{O}_{2}}\] at NTP So, the volume strength of \[1.5\text{ }N\text{ }{{\text{H}}_{2}}{{O}_{2}}\]solution \[=8.4\]volume.
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