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A 500 mL sample of '45.4 volume' H2O2 solution is left exposed to the atmosphere and loses 11.2 litre of O2 measured at 1 atm and 273 K. Assuming the volume of solution stays constant, what is the new molarity of the H2O2 solution?
- 2.0 M
- 0.5 M
- 1.0 M
- 1.5 M
Correct answer: 2.0 M
Solution
Initial molarity = 45.4/11.2 ~ 4.054 M giving 2.027 mol in 500 mL; 11.2 L O2 at STP = 0.5 mol O2 consuming 1.0 mol H2O2, leaving ~1.0 mol in 0.5 L = 2.0 M.
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