Question

Calculate the molality of 1 litre solution containing 93\% $$$${H_2}S{O_4} $\left( {W/V} \right)$ if the density of the solution is 1.84 $g\,m{l^{ - 1}}$

Answer 1

Molality is the ratio of the number of moles of the solute to the total mass of the solvent. To calculate molality we need to first find the mass of ${H_2}S{O_4}$ and then calculate the molality by given formula.

Formula used: Molality = $\dfrac{{no.\,of\,moles\,of\,solute}}{{mass\,of\,solvent}}$
Density $= \dfrac{{mass}}{{volume}}$

Complete Step-by-Step Answer:
Before we move towards the solution of this question, let us discuss some basic concepts.
In the given question, a solution of $93\%$ {H_2}S{O_4}$$$$\left( {W/V} \right) is present. This means that 93 g of ${H_2}S{O_4}$ is present in 100 ml of the solution. Hence, in 1 litre of solution, we will have 930 g of ${H_2}S{O_4}$. In addition to that, the density of the solution is given as 1.84 $g\,m{l^{ - 1}}$. Also, the molar mass of solute, i.e. ${H_2}S{O_4}$ is 98$g/mol$. Hence, the total mass of the solute present can be calculated as:
Density $= \dfrac{{mass}}{{volume}}$
Mass = (density) (volume) = $\left( {1.84} \right){\text{ }}\left( {1000} \right){\text{ }} = {\text{ }}1840$ g of ${H_2}S{O_4}$
Now,
Molality = $\dfrac{{no.\,of\,moles\,of\,solute}}{{mass\,of\,solvent}}$= $\dfrac{{(930)\,(1000)}}{{(98)\,(910)}}$= 10.43 $mol/kg$

Note: Molality and molarity are two different concepts. Students usually get confused between these two. Hence,
Molarity can be defined as the amount of substance present in a given volume of a solution. In mathematical terms, it is the ratio of the number of moles of the solute to the volume of the solution. Representing this in the form of an equation:
Molarity = $\dfrac{{no.\,of\,moles\,of\,solute}}{{volume\,of\,solution\,(in\,Litres)}}$