The correct options are
B 15 mL of the same sample of H2O2 solution liberates 224 mL of O2 gas at 1.5 atm and 273oC
C Milliequivalents of hypo required for the titration of liberated I2 when 10 mL of the same sample of H2O2 solution is treated with excess of acidified solution of KI are 20
D % (w/v) of given same sample of H2O2 solution is 3.4%
10 mL of H2O2 solution liberates 112 mL of O2 at STP.
Therefore,
1 L of H2O2 solution liberates 11.2 L O2 at STP
Hence,
Volume strength of H2O2 solution = 11.2
Normality=11.25.6=2
% (w/v)=N×1710=3.4%
15 mL of same sample of H2O2 solution liberates 15×11210 mL O2 at STP
⟹15×11210×23×2=224 mL O2 at 1.5 atm and 273 oC.
milliequivalents sample of H2O2 reacted with KI
= N× V
= 2× 10
= 20
milliequivalents of hypo required = 20