Question

The volume strength 1 M of ${ H }_{ 2 }{ O }_{ 2 }$ is
(Molar mass of ${ H }_{ 2 }{ O }_{ 2 }$ = 34g mol/1)
a.) 16.8
b.) 11.2
c.) 22.4
d.) 5.6

Answer 1

Hint : Volume strength of ${ H }_{ 2 }{ O }_{ 2 }$ is the volume of ${ O }_{ 2 }$ released on decomposition of one volume of ${ H }_{ 2 }{ O }_{ 2 }$ at STP. With the reaction of decomposition of hydrogen peroxide we will calculate volume strength of 1M solution of hydrogen peroxide.

Complete step by step solution :
Volume or percentage strength of hydrogen peroxide is a term to define concentration of ${ H }_{ 2 }{ O }_{ 2 }$ in terms of volume of Oxygen gas based on its decomposition to form water and oxygen.

${H }_{ 2 }{ O }_{ 2 }\longrightarrow { H }_{ 2 }O\quad +\quad { O }_{ 2 }$

So, for finding volume strength we have to write a balanced reaction of ${ H }_{ 2 }{ O }_{ 2 }$ to decomposition to form water and oxygen i.e.

$2{H }_{ 2 }{ O }_{ 2 }\longrightarrow 2{ H }_{ 2 }O\quad +\quad { O }_{ 2 }$

Here 2 moles of ${{H}_{2}}{{O}_{2}}$ decomposes to give 2 moles of water and 1 mole of oxygen gas at STP (standard temperature=273K and pressure is 1atm).
As we know according to Avagadro’s law, the ratio of volume and amount of gaseous substance is a constant. The value of this constant (k) can be measured by the following equation:-
$k=\dfrac{RT}{P}$ R= gas constant
At standard temperature and pressure, the value of T (temperature) is 273.15 Kelvin and the value of P (pressure) corresponds to 101.325 kiloPascals (1atm is equal to 101.325 kilopascal). Hence, the volume occupied by one mole of a gas will be:-

Volume occupied by 1 mole of gas=$\dfrac{\left( 8.314 J mo{{l}^{-1}}{{K}^{-1}} \right)\left( 273.15K \right)}{101.325 kPa}=22.4L$

So,in terms of volume, 2 moles ${{H}_{2}}{{O}_{2}}$ at STP gives 22.4 L ${{O}_{2}}$
And thus, 1 mole ${{H}_{2}}{{O}_{2}}$ gives 11.2 L ${{O}_{2}}$
. So, from the reaction stoichiometry-

1 mole of ${{H}_{2}}{{O}_{2}}$ =11.2L of ${{O}_{2}}$.

By using the formula of volume strength i.e.
Volume strength of ${{H}_{2}}{{O}_{2}}$ = $Molarity\times volume\text{ }of\text{ }oxygen$
Where M is the molarity which is defined as the number of moles of solute per litre of volume of solution.
$\begin{aligned} & volume\quad strength\quad =\quad 1\quad \times \quad 11.2\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\\ & \quad \quad \quad \quad \quad \quad \quad \quad =\quad 11.2\quad volume \\\ \end{aligned}$
So, the correct answer is “Option B”.

Additional Information:
The oxidation state of Oxygen in ${{H}_{2}}{{O}_{2}}$ is ‘-1’ (peroxide) whereas the oxidation state of oxygen in water is ‘-2’ (oxide).

Note : Balancing an equation accurately is most important. Don’t confuse molecular mass with molar mass as the unit of molar mass is given in g/mol because molar mass is slightly different from molecular mass. As molecular mass is the mass of one molecule of a compound and its unit is gm only while molar mass is the mass of one mole of a compound thus its unit is gm/mol.