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# Engineering

The strength of 11.2 volume solution of H_{2}O_{2} is [ Given that molar mass of H=1 g mol-1 and O=16 g mol-1]

  • Option 1)

    13.6%

  • Option 2)

    3.4%

  • Option 3)

    34%

  • Option 4)

    1.7%

Answers (1)
P Plabita

 

volume strength of H2O2 -

Volume or percentage strength of hydrogen peroxide is a term to express concentration of H2O2 in terms of volumes of oxygen gas ,based on its decomposition to form water and oxygen.

2H2O2 = 2 H2O + O2

Suppose we have a solution of M moles of H2O2 in 1L solution

From the reaction stoichiometry, 2 moles of H2O2 gives 1mol of oxygen or 22.4L at S.T.P

1mole of H2O2 = 11.2 L of O2

M mole of H2O2 = 11.2 x M L of O2

1L of H2O2 gives =11.2 x M L of O2

Vol strength of H2O2 = 11.2 x M

In terms of Normality, N

n factor for H2O2 is 2

N = M x n or N = M x 2

V S = 11.2 M = 11.2 x N/2 = 5.6 Na

-

 

 

 

Let:- 2H_{2}O_{2}(aq)\rightarrow 2H_{2}O+O_{2}(g)

\Rightarrow Volume strength of H2O2 refers to the Number of volumes of oxygen gas evolved by one volume of H2O2

\Rightarrow Acc. to equation

2 moles of H2O2=22.4 L of oxygen at STP

That means

2 \times 34g \;H_{2}O_{2}=22.4 L O_{2}

68g \;H_{2}O_{2}=22.4 L O_{2}

OR

Mole of O2 gas formed= \frac{1}{2}

mole of H2O2 present in 1 litre solution=1

mass of H2O2  in 1 litre solution= 1X34=34 gm

% \frac{N}{V} strength= \frac{34 gm}{1000 ml}\times 100=3.4 %


Option 1)

13.6%

Option 2)

3.4%

Option 3)

34%

Option 4)

1.7%

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