Question

What is the molarity of $11.2$ $V$ of {H_2}{O_2}$$$$?
A.$1$ M
B.$2$ M
C.$5.6$ M
D.$11.2$ M

Answer 1

First we know that the molarity (M) of a solution is the number of moles of solute dissolved in one litre of solution. The solution includes both the solute and the solvent. Hence the molarity of a solution is the ratio of the moles of solute to the volume of the solution expressed in litres.

Complete answer:
Given $11.2$ $V$ of ${H_2}{O_2}$ which is the solute. Let $n$be the number of moles of solute ${H_2}{O_2}$ in a solution and $v$be the volume of a solution.
We know that one litre of ${H_2}{O_2}$​ aqueous solution provides $11.2$ litre of ${O_2}$​ at STP (Standard temperature and pressure).
Moles of {O_2}$$$$ = \dfrac{{11.2}}{{22.4}} = 0.5
Then the number of moles in ${H_2}{O_2}$ is $n = 0.5 \times 2$.
molarity of${H_2}{O_2}$(M)​​ $= \dfrac{n}{v} = 1$M
Hence the correct option is (A) $1$M.

Note:
We can also solve this problem by using the relationship between normality and molarity.
i.e., $Molarity = \dfrac{{Normality}}{n}$---(1)
We know that one normality ($N$) of ${H_2}{O_2}$ is $5.6$ times the volume strength ($V$).
Then $1\;V = \dfrac{1}{{5.6}}N$
$\Rightarrow 11.2\;V = \dfrac{{11.2}}{{5.6}}N = 2N$
We know that $n = 2$ for ${H_2}{O_2}$
Then the equation (1) becomes
$Molarity = \dfrac{2}{2} = 1M$
Also note that the volume is in litres of solution and not litres of solvent. At Standard temperature and pressure, one mole of any gas will occupy a volume of $22.4$ litre.