Question

The mole fraction of hydrogen peroxide in 30% (weight/weight) aqueous solution of ${H_2}{O_2}$ is ______________________.

Answer 1

A mole, as per International System of Units is the measure of the amount of substance, and one mole is equal to $6.023 \times {10^{23}}$ and mole fraction of a particle is the ratio of one component to the total number of moles of the solution.

Complete step by step answer:
Assume the total weight of the hydrogen peroxide and water is 100g. As it is given that the weight of Hydrogen peroxide is 30, then the weight of water is calculated as:
$100 - weight{\text{ }}of{\text{ }}hydrogen$
$\Rightarrow 100 - 30$
$\Rightarrow 70g$
As per definition, a mole fraction is defined as the ratio of the number of moles of one component to the total number of moles of solution. And the number of moles of any component is calculated as the ratio of the given weight of that component to the molecular weight of that component.
Since the molecular weight of hydrogen peroxide is equal to the sum of all the hydrogen and oxygen atoms present in it. The molecular weight of hydrogen peroxide is:
$2 \times 1 + 2 \times 16 = 34g$
And the molecular weight of the water is:
$2 \times 1 + 16 = 18g$
The number of moles of hydrogen peroxide is given as:
${n_{{H_2}{O_2}}} = \dfrac{{{\text{given mass of hydrogen peroxide}}}}{{{\text{molecular mass of hydrogen peroxide}}}}$
${n_{{H_2}{O_2}}} = \dfrac{{30}}{{34}} = 0.8823$mole
${n_{{H_2}O}} = \dfrac{{{\text{given mass of water}}}}{{{\text{molecular mass of water}}}}$
${n_{{H_2}O}} = \dfrac{{70}}{{18}} = 3.88$mole
Mole fraction of hydrogen peroxide = $\dfrac{{{n_{{H_2}{O_2}}}}}{{{n_{{H_2}{O_2}}} + {n_{{H_2}O}}}}$
$\Rightarrow \dfrac{{0.8823}}{{3.88 + 0.8823}}$
$\Rightarrow \dfrac{{0.8823}}{{4.76}} = 0.184$

Thus the mole fraction of hydrogen peroxide in a 30% aqueous solution of hydrogen peroxide is 0.184.

Note: The sum of the mole fraction of both components .i.e. solute and solvent are always equal to 1. This is shown as:
Let's assume there is a solution AB in which A represents solute whereas B represents solvent. The mole of solute A is given as:
${x_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}$ (1)
Similarly, the mole fraction of solvent B is given as:

${x_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}$ (2)
Add equation (1) and (2), we get:
${x_A} + {x_B} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}} + \dfrac{{{n_B}}}{{{n_A} + {n_B}}}$
On taking L.C.M we get:
$\Rightarrow \dfrac{{{n_B} + {n_A}}}{{{n_A} + {n_B}}} = 1$
Hence proved. And also mole fraction is a unitless quantity because it is the ratio of the same quantity.