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Grade: 12th pass

                        

The normality of 10 mL of a ‘20V’ H 2 O 2 solution is? [AC ch5 pg 31 Q34]

The normality of 10 mL of a ‘20V’ H2O2 solution is?
 
 
 
[AC ch5 pg 31 Q34]

6 years ago

Answers : (2)

Naveen Kumar
askIITians Faculty
60 Points
							20VH2O2 means that 1 litre of H2O2 solution would produce 20 litre of Oxygen.

2H2O2.....................>H2O+ O2

From the equation it is clear that 2 moles of H2O2 produce 1 mol of O2
But 20V H2O2 will produce 20 litre of oxygen for 1 litre of H2O2.
............................................⇒ (20litre/24.4litre) mol of oxygen liberated for 1 litre of H2O2
So in 1 litre solution of H2O2, moles of H2O2=2*moles of oxygen = (2*20/24.4) mol/litre=1.67M
n-factor of H2O2=2
So, normality=Molarity*n-factor=1.67*2=3.34N

6 years ago
Kushagra Madhukar
askIITians Faculty
629 Points
							
Dear student,
Please find the attached solution to your problem below.
 
20V H2O2 means that 1 litre of H2O2 solution would produce 20 litre of Oxygen.
2H2O2 → H2O + O2
From the equation it is clear that 2 moles of H2O2 produce 1 mol of O2
But 20V H2O2 will produce 20 litre of oxygen for 1 litre of H2O2.
⇒ (20litre/24.4litre) mol of oxygen liberated for 1 litre of H2O2
So in 1 litre solution of H2O2, moles of H2O2 = 2*moles of oxygen = (2*20/24.4) mol/litre = 1.67M
n-factor of H2O2 = 2
So, normality = Molarity*n-factor = 1.67*2 = 3.34N
 
Hope it helps.
Thanks and regards,
Kushagra
7 months ago
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