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Question: 2 liters of sample of \({{H}_{2}}{{O}_{2}}\) give 22.4 liters of \({{O}_{2}}\) at S.T.P. Normality o...

2 liters of sample of H2O2{{H}_{2}}{{O}_{2}} give 22.4 liters of O2{{O}_{2}} at S.T.P. Normality of given H2O2{{H}_{2}}{{O}_{2}} sample is ‘X’.

Explanation

Solution

First we have to find out the strength of the given H2O2{{H}_{2}}{{O}_{2}} sample by apply Avogadro’s law which states that, one of mole of gas occupies 22.4 liters of volume at S.T.P and N.T.P and by using the normality formula of H2O2{{H}_{2}}{{O}_{2}} as ;Normality=Volume strength5.6Normality={Volume\text{ }strength}{5.6}, we can easily calculate its normality. Now solve it.

Complete step by step solution:
- First of all, let’s discuss what is normality. By the term normality , we mean the total no of moles of gram equivalent dissolved in 1 litre of the solution and the normality is represented by the symbol as N. The normality of H2O2{{H}_{2}}{{O}_{2}} can be directly found by using the formula as:
Normality=Volume strength×5.6Normality={Volume\text{ }strength} \times {5.6}
- This numerical is based on the Avogadro’s law which states that the one mole of the gas occupies 22.4 liters of volume at S.T.P. or N.T.P
Now considering the numerical;
2 liters of sample of H2O2{{H}_{2}}{{O}_{2}} give = 22.4 liters22.4\text{ }liters of O2{{O}_{2}} at S.T.P. ( given)
Then,
1 liters of sample of H2O2{{H}_{2}}{{O}_{2}} give =22.42 liters{22.4}{2}\text{ }liters of O2{{O}_{2}} at S.T.P.
= 11.2 liters11.2\text{ }liters of O2{{O}_{2}} at S.T.P.
Now, by applying the normality formula for H2O2{{H}_{2}}{{O}_{2}},we can easily find its normality as;
Normality=Volume strength×5.6Normality={Volume\text{ }strength} \times {5.6}-----------(1)
Volume strength =11.4 liters11.4\text{ }liters of O2{{O}_{2}} at S.T.P.
Normality = X (given)
Put all these values in equation (1), we get;
X=11.4×5.6  = 2 N \begin{aligned} & X={11.4}\times{5.6} \\\ & \text{ = 2 N} \\\ \end{aligned}

Therefore, 2 liters of sample of H2O2{{H}_{2}}{{O}_{2}} which gives 22.4 liters of O2{{O}_{2}} at S.T.P. has the Normality as 2 N2\text{ }N.

Note: Avogadro’s law is only applicable under standard conditions of temperature and pressure i.e. at 1 atmosphere pressure and 273 K temperature and if these conditions are not satisfied , then this law is not applicable in those conditions.