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.)

### 1 Answer | Add Yours

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,

1*5.6=v*2

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.

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes