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Question

If 10 ml of an aqueous solution of H2O2 liberates 112 ml of O2 gas at STP upon decomposition, then:

A
normality of the above sample of H2O2 is 0.5N
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B
molarity of above sample of H2O2 is 1M
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C
milliequivalents of hypo required for the titration of liberated I2 when 10 ml of the same sample of H2O2 is treated with excess of acidified solution of KI is 20
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D
%(w/v) of given sample of H2O2 is 3.4%
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Solution

The correct options are
B %(w/v) of given sample of H2O2 is 3.4%
C molarity of above sample of H2O2 is 1M
D milliequivalents of hypo required for the titration of liberated I2 when 10 ml of the same sample of H2O2 is treated with excess of acidified solution of KI is 20
The reaction is as follows:
H2O2H2O+12O2
Volume of 1 mmol gas at STP=22.4 ml.
No. of moles in 112 ml of O2=11222.4=5 mmol.
Hence, according to reaction,
Number of moles of H2O2 =2×5 mmol =10 mmol.
Hence, molarity=Number of MolesVolume of solution =10mmol10ml=1M

Milli equivalent of H2O2 reacted with KI=N×V=2×10=20
Milli equivalent of hypo required =20
%(w/v) calculation
Molar weight of H2O2=34 g
Weight of 10 mmol of H2O2=10×103×34=0.34 g
%(w/v) =0.34gm10ml×100=3.4%

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