Question

Volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ solution has percent strength of about
(A) 4.55
(B) 5.44
(C) 3.12
(D) 1.56

Answer 1

Hint: We know that the amount of the solute or the concentration of the solute that is present in the total solution is known as the strength of the solution. We are given 15 volumes of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ where volume strength indicates the volume of oxygen produced by one litre of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$.

Formulae Used:
(1 )${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume}}}}$

(2 )${\text{Strength}} = {\text{Molarity}} \times {\text{Molar mass}}$

(3) $\% {\text{w/w}} = \dfrac{{{\text{Mass of solute}}}}{{{\text{Mass of solution}}}} \times 100$

Complete Answer:
We know that volume strength is the number of volumes of oxygen liberated by one litre of hydrogen peroxide.
We know that two moles of hydrogen peroxide produce two moles of water and one mole of oxygen. The reaction is as follows:
${\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} \to {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}} + {{\text{O}}_{\text{2}}}$
From the reaction, we can see that two moles of hydrogen peroxide produce two moles of water and one mole of oxygen.
We will first calculate the molar mass of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ as follows:
Molar mass of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ \[ = \left( {{\text{2}} \times {\text{Molar mass of H}}} \right) + \left( {2 \times {\text{Molar mass of O}}} \right)\]
Molar mass of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ \[ = \left( {{\text{2}} \times 1} \right) + \left( {2 \times {\text{16}}} \right)\]
Molar mass of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ \[ = 34{\text{ g/mol}}\]
Thus, the molar mass of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is \[34{\text{ g/mol}}\].

Thus, one mole of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ contains \[34{\text{ g}}\] of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$. Thus, two moles of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ contains \[68{\text{ g}}\] of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$.
We know that one mole of an ideal gas at standard temperature and pressure occupies $22.4{\text{ L}}$ of volume.
Assume that the volume strength is x. Thus, x litre of oxygen is produced from $\dfrac{{68x}}{{22.4}}{\text{g}}$ of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$.
Thus, the volume strength for x litre of oxygen is $\dfrac{{68x}}{{22.4}}$.

Now, calculate the molarity of the solution using the equation as follows:
${\text{Molarity}} = \dfrac{{{\text{Number of moles}}}}{{{\text{Volume}}}}$ …… (1)
We know that the number of moles is the ratio of mass to the molar mass. Thus, equation (1) becomes,
${\text{Molarity}} = \dfrac{{\dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}}}{{{\text{Volume}}}}$
Substitute $\dfrac{{68x}}{{22.4}}{\text{g}}$ for the mass, \[34{\text{ g/mol}}\] for the molar mass and $1{\text{ L}}$ for the volume.
Thus,
${\text{Molarity}} = \dfrac{{\dfrac{{\dfrac{{68x}}{{22.4}}{\text{g}}}}{{34{\text{ g/mol}}}}}}{{1{\text{ L}}}}$
${\text{Molarity}} = \dfrac{x}{{11.2}}$
We are given 15 volumes of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$.
Thus,
${\text{Molarity}} = \dfrac{{15}}{{11.2}} = 1.34$
Thus, the molarity of the solution is $1.34{\text{ M}}$.

Now, calculate the strength using the equation as follows:
${\text{Strength}} = {\text{Molarity}} \times {\text{Molar mass}}$
${\text{Strength}} = {\text{1}}{\text{.34}} \times {\text{34}}$
${\text{Strength}} = 45.5{\text{ g/L}}$
Thus, the strength of the solution is $45.5{\text{ g/L}}$.

Thus, the weight ratio is,
$\% {\text{w/w}} = \dfrac{{{\text{Mass of solute}}}}{{{\text{Mass of solution}}}} \times 100$
$\% {\text{w/w}} = \dfrac{{{\text{34}}}}{{{\text{1000}}}} \times 100$
$\% {\text{w/w}} = 4.55\% $
Thus, 15 volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ solution has percent strength of about $4.55\% $.

Thus, the correct option is (A) $4.55\% $.

Note: Hydrogen peroxide decomposes very easily. Hydrogen peroxide easily decomposes to two components i.e. water and oxygen.Molarity and molality should not be confused both are different molarity is temperature dependent while molality is temperature independent. Unit conversions in such questions should be done carefully.