Question

Molarity of aqueous ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $3{\text{ M}}$. If the density of solution is $2{\text{ gm/ml}}$ find molality, strength, normality and $\% {\text{ wt/wt}}$ and mole fraction.

Answer 1

Hint: Molarity, molality, strength, normality, $\% {\text{ wt/wt}}$ and mole fraction are all used to express the concentration of solutions. We are given the molarity of aqueous ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$. The number of moles of a solute dissolved in a liter of solution is known as its molarity. To solve this we must know the relations between all these quantities.

Complete solution:
We are given the molarity of aqueous ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$. The molarity of aqueous ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $3{\text{ M}}$.
We know that the number of moles of a solute dissolved in a liter of solution is known as its molarity. Thus, $1000{\text{ ml}}$ solution contains $3{\text{ mol}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$.
Now, we know that number of moles is the ratio of mass to the molar mass. The molar mass of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $98{\text{ g/mol}}$. Thus,
${\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} = 3{\text{ mol}} \times {\text{98 g/mol}} = 294{\text{ g}}$
We know that density is the ratio of mass to the volume. The expression for density is as follows:
$d = \dfrac{m}{V}$
Thus,
$m = d \times V$
Substitute $2{\text{ gm/ml}}$ for the density, $1000{\text{ mL}}$ for the volume of the solution. Thus,
$m = 2{\text{ g/mL}} \times 1000{\text{ mL}}$
$m = 2000{\text{ g}}$
Thus, the mass of the solution is $2000{\text{ g}}$.
Now, the mass of water i.e. solvent is the difference in the masses of the solution and ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$. Thus,
${\text{Mass of water}} = \left( {2000 - 294} \right){\text{ g}} = 1706{\text{ g}}$
Thus, the mass of water i.e. solvent is $1706{\text{ g}} = 1706 \times {10^{ - 3}}{\text{ kg}}$.
We know that the molality of a solution is the number of moles of solute per kilogram of solvent. Thus,
${\text{Molality}} = \dfrac{{3{\text{ mol}}}}{{{\text{1706}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ kg}}}} = 1.75{\text{ m}}$
Thus, the molality of aqueous ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $1.75{\text{ m}}$.
Strength of any solution is the concentration of the solution in ${\text{g/L}}$.
We know that $294{\text{ g}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is present in the solution.
Thus, the strength of aqueous ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $294{\text{ g/L}}$.
The normality of a solution is defined as the number of grams equivalent of solute per liter of solution.
The equation to calculate normality is,
${\text{Normality}} = n \times {\text{Molarity}}$
 $n$ is the number of hydrogen atoms in ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$.
${\text{Normality}} = 2 \times {\text{3 M}} = 6{\text{ N}}$
Thus, the normality of aqueous ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $6{\text{ N}}$.
Weight by weight percentage can be calculated by dividing the mass of solute by the mass of solution and then multiplying it by 100. Thus,
$\% {\text{ wt/wt}} = \dfrac{{294{\text{ g}}}}{{2000{\text{ g}}}} \times 100 = 14.7{\text{\% }}$
Thus, the $\% {\text{ wt/wt}}$ of aqueous ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is $14.7{\text{\% }}$.
Mole fraction of any component is the ratio of moles of any one component to the total number of moles.
Now, we have $1706{\text{ g}}$ of water. The molar mass of water is $18{\text{ g/mol}}$. Thus,
${\text{Number of moles of water}} = \dfrac{{1706{\text{ g}}}}{{18{\text{ g/mol}}}} = 94.77{\text{ mol}}$
Mole fraction of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ $ = \dfrac{{3{\text{ mol}}}}{{94.77{\text{ mol}} + 3{\text{ mol}}}} = 0.0306$
Mole fraction of water $ = \dfrac{{94.77{\text{ mol}}}}{{94.77{\text{ mol}} + 3{\text{ mol}}}} = 0.969$

Thus, mole fraction of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is 0.0306 and mole fraction of water is 0.969.

Note:Remember all the terms. Molarity is the number of moles of solute per litre of solution. Molality is the number of moles of solute per kilogram of solvent. Strength is the concentration of solution in ${\text{g/L}}$. Normality of a solution is the number of gram equivalent of solute per liter of solution. Weight by weight percentage can be calculated by dividing the mass of solute by the mass of solution and then multiplying it by 100. And the mole fraction of any component is the ratio of moles of any one component to the total number of moles.