Question

Calculate the molality of $1{\text{ L}}$ solution of $93\% {\text{ }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ (weight/volume). The density of the solution is $1.84{\text{ g/mL}}$.
(If the answer is x, then find $\left[ x \right]$)

Answer 1

To solve this we must know the term molality. The number of moles of a solute per kilogram of a solvent is known as molality. First calculate the mass of the solvent. From the mass of the solvent and the number of moles of solute calculate the molality of the solution.

Formulae Used:

1. $d = \dfrac{m}{V}$
2. ${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
3. ${\text{Molality}}\left( {\text{m}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Mass of solvent}}\left( {{\text{kg}}} \right)}}$

Complete step by step solution:
We know that density is the ratio of mass to the volume. The expression for density is as follows:
$d = \dfrac{m}{V}$
Where $d$ is the density of the solution,
$m$ is the mass of the solution,
$V$ is the volume of the solution,
Rearrange the equation for the mass of the solution as follows:
$m = d \times V$
Substitute $1.84{\text{ g/mL}}$ for the density, $1{\text{ L}} = 1000{\text{ mL}}$ for the volume of the solution. Thus,
$m = 1.84{\text{ g/mL}} \times 1000{\text{ mL}}$
$m = 1840{\text{ g}}$
Thus, the mass of the solution is $1840{\text{ g}}$.
We are given $93\% {\text{ }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ (weight/volume). $93\% {\text{ }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ (weight/volume) means that $93{\text{ g }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is present in $100{\text{ mL}}$ of water.
If $93{\text{ g }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is present in $100{\text{ mL}}$ of water then $930{\text{ g }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is present in $1000{\text{ mL}}$ of water.
Thus, the mass of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is $930{\text{ g}}$.
Now, the mass of water is the difference in the masses of the solution and ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$. Thus,
${\text{Mass of water}} = \left( {1840 - 930} \right){\text{ g}}$
${\text{Mass of water}} = 910{\text{ g}}$
Thus, the mass of water i.e. solvent is $910{\text{ g}} = 910 \times {10^{ - 3}}{\text{ kg}}$.
We know that the number of moles is the ratio of mass to the molar mass. Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Calculate the number of moles of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ as follows:
Substitute $930{\text{ g}}$ for the mass of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$, $98{\text{ g/mol}}$ for the molar mass of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$. Thus,
${\text{Number of moles of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = \dfrac{{930{\text{ g}}}}{{98{\text{ g/mol}}}}$
${\text{Number of moles of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = 9.49{\text{ mol}}$
Thus, the number of moles of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ i.e. solute are $9.49{\text{ mol}}$.
We know that the molality of a solution is the number of moles of a solute per kilogram of a solvent. Thus,
${\text{Molality}}\left( {\text{m}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Mass of solvent}}\left( {{\text{kg}}} \right)}}$
Substitute $9.49{\text{ mol}}$ for the number of moles of solute, $910 \times {10^{ - 3}}{\text{ kg}}$ for the mass of solvent. Thus,
${\text{Molality}} = \dfrac{{9.49{\text{ mol}}}}{{910 \times {{10}^{ - 3}}{\text{ kg}}}}$
${\text{Molality}} = 10.4{\text{ m}}$
Thus, the molality of $1{\text{ L}}$ solution of $93\% {\text{ }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ (weight/volume) is $10.4{\text{ m}}$.

Thus, $x = 10.4{\text{ m}}$.

Note: The molality of a solution is the number of moles of a solute per kilogram of a solvent. Do not confuse between the mass of solution and the mass of solvent. Here the solvent is water and the solution is a mixture of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ and water. Thus, we have to consider the mass of water and not the solution.