Question

Find molarity, molality and normality of ${H_2}S{O_4}$ in $98\%$ w/w solution. (Density $\rho = 1.8gm{l^{ - 1}}$ )

Answer 1

Molarity (M), molality (m) and normality (N) are units of concentration. Molarity $\left( M \right)$ is the number of moles of solute present in $1L$ of solution. Molality (m) is the number of moles of solute present in $1kg$ of solvent. Normality (N) is the number of gram equivalents of solute present in $1L$ of solution.

Complete answer:
We have a $98\%$ w/w solution which means $98g$ of solute present in $100g$ of solution. Also, the density given is $1.8gm{l^{ - 1}}$
Firstly we’ll find molarity;
Molar mass of ${H_2}S{O_4}$ = $98gmo{l^{ - 1}}$
Amount of ${H_2}S{O_4}$ in solution = $98g$
Moles of ${H_2}S{O_4}$ = $\dfrac{{98}}{{98}}$
$= 1$ mole
We know that $Density = \dfrac{{mass}}{{volume}}$
We can find the volume of solution from here;
$Volume = \dfrac{{100}}{{1.8}}$ ml
$\dfrac{{100}}{{1.8}}ml$ of solution $\to 1$ mole
$1000ml(1L)$ of solution $\to \dfrac{{1 \times 1000 \times 1.8}}{{100}}$ moles
$= 18$ moles
Thus, the molarity (No of moles of solute present in 1L of solution) was found to be 18M
Now, Molality;
It is given that $98g$ of solute is present in $100g$ of solution. (Solvent = $100 - 98$ )
$2g$ of solvent $\to 1$ mole
$1000g$ ( $1kg$ ) solvent $\to \dfrac{{1 \times 1000}}{2}$
$= 500$ moles
Thus, the molality (no of moles of solute present in $1kg$ of solvent ) was found to be $500m$
For Normality;
There is a formula present that relates molarity and normality. We can use that formula to find out normality.
$Normality = n - factor \times Molarity$
For ${H_2}S{O_4}$ n-factor is $2$
Normality = $2 \times 18$
= $36N$
Thus, the normality (No of gram equivalents of solute present in $1L$ of solution) was found to be $36N$

Note:
We have seen how to find molarity, molality and normality when density is given. There are direct formulas also available to find the same if density and w/w percent is given. Here we have used conventional methods to find the concentrations.