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

Hint: 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 = \[\dfrac{{no.\,of\,moles\,of\,solute}}{{volume\,of\,solution\,(in\,Litres)}}\]