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Q.Which of the following is/are correct for 17 g/L of H2O2 SOLUTION? A) Volume...
Q.Which of the following is/are correct for 17 g/L of H2O2 SOLUTION?
A) Volume strength is 5.6 at 273 K and 1 atm.
B) Molarity of solution is 0.5 M.
C) 1 mL of this solution gives 2.8 mL O2 at 273 K and 2 atm.
D) The normality of solution is 2 N.
(Please mark the correct answers with proper explanations.And also show the correct answer in the work.)
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Best answer as selected by question asker.
Volume strength of H2O2 is defined as the number of volumes of oxygen that is released on decomposition of 1 volume of a particular hydrogen peroxide solution. The decomposition follows equation:
2H2O2 (aq.) → 2H2O + O2 (g)
It follows that 68 g H2O2 produces 22400 ml O2 at STP
Therefore, 17 g H2O2 will produce 22400*68\17 = 5600 ml O2 at STP.
When 17 g H2O2 is present in 1000 ml solution, we can write
1000 ml aq. H2O2 solution produces 5600 ml O2 at STP (i.e. at 273 K and 1 atm.)
So, 1 ml aq. H2O2 solution produces 5600/1000 = 5.6 ml O2 at STP
Volume strength of 17 g/L aq. H2O2 solution is thus 5.6 volume. Option A) is correct.
Molarity of H2O2 is defined as number of moles present in 1 L solution.
Molar mass of H2O2 is 34 g
Hence 17 g/L aq. H2O2 solution is 17/34 M = 0.5 M. Option B) is correct.
Normality of H2O2 is defined as the number of gm-equivalents present in 1 L solution.
Equivalent mass of H2O2 is the mass that contains 1 g hydrogen, i.e. 34/2 = 17 g
Hence 17 g/L aq. H2O2 solution is 17/17 M = 1.0 N. Option D) is wrong.
5.6 ml O2 at STP = v ml O2 (say) at 273 K and 2 atm. pressure.
Applying gas laws,
rArr v = 5.6/2 =2.8 ml
Thus 1 ml aq. H2O2 solution produces 5.6 ml O2 at STP =2.8 ml O2 at 273 K and 2 atm. Pressure.
Option C) is also correct.
Therefore, the options A), B) and C) are correct.
Posted by llltkl on June 25, 2013 at 4:26 PM (Answer #1)
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