Question

Molarity of ${H_2}S{O_4}$ is $0.8M$ and its density $1.06\,g/c{m^3}$. What will be the concentration of the solution in terms of molality and molar fraction?

Answer 1

Molarity, Molality and mole fraction are the widely used units for determining the concentration of solutions. Here, molarity is given which is the ratio of solvent's moles to the solutions total volume in liters through which helps in determining the number of moles of the solution.

Complete answer:
Let’s understand the step by step solution of this question:
Here, Molarity of ${H_2}S{O_4}$=$0.8M$= $0.8\,mol\,{L^{ - 1}}$
This indicates that $0.8\,mol\,$is present in one liter of the solution.
Also, Number of moles of ${H_2}S{O_4}$ = $\dfrac{{Given\,\,mass}}{{Molar\,mass}}$
Mass of${H_2}S{O_4}$ =$Moles\,\,of\,{H_2}S{O_4} \times \,Molar\,mass$
Mass of ${H_2}S{O_4}$= $\,0.8\,mol \times \,98\,g\,mo{l^{ - 1}}$
Mass of ${H_2}S{O_4}$= $78.4g$
Mass of $1$L of solution=$1000\,c{m^3}\,\, \times \,\,1.06\,g\,c{m^3}\, = 1060g$
Mass of water in solution=$1060g\, - 78.4g\, = 981.6g = 0.9816Kg$
Molality (m) = $\dfrac{{Number\,of\,moles\,of\,{H_2}S{O_4}}}{{Mass\,of\,solvent\,\,in\,\,Kg}}$
Molality (m) =$\dfrac{{0.8}}{{0.9816\,}}\, = 0.815\,m$
Now, we calculate the mole fraction of ${H_2}S{O_4}$. For this, we need to determine the number of moles of ${H_2}O$
Number of moles of ${H_2}O$= $\dfrac{{Given\,Mass}}{{Molar\,Mass}}$
Number of moles of ${H_2}O$= $\dfrac{{981.6}}{{18}} = \,54.5\,mol$
Put the values of number of moles of ${H_2}O$and ${H_2}S{O_4}$in mole fraction formula i.e.
Mole fraction of ${H_2}S{O_4}$= $\dfrac{{{n_{{H_2}S{O_4}}}}}{{{n_{{H_2}S{O_4}}}\, + \,\,{n_{{H_2}O}}}}$
Mole fraction of ${H_2}S{O_4}$=$\dfrac{{0.8}}{{0.8\,\, + \,54.5}} = \,\dfrac{{0.8}}{{55.3}}$
Mole fraction of ${H_2}S{O_4}$= $0.014$
Therefore, the concentration of ${H_2}S{O_4}$ in terms of molality is $0.815\,mol$ and its mole fraction is $0.014$.

Note:
Do not confuse between molarity and molality. Molarity is related with the volume of solution in liters whereas, molality is related with the mass of the solvent in Kg. Also, remember the units of molarity i.e. $mol\,{L^{ - 1}}$and molality i.e. $mol\,K{g^{ - 1}}$. Molar fraction is just the ratio of the number of moles so, it does not have any units.