Question

What is the molarity of 0.2N ${H_2}S{O_4}$ ?

Answer 1

Hint: Molarity can be defined as the number of moles of solute present in one litre of solution. Normality can be defined as the number of gram equivalents of solute present in one litre of solution. Molarity and normality both are units of concentration.

Complete answer:
We have seen that both molarity and normality are units of concentration.
Now, coming to our question;
We have $0.2N$ ${H_2}S{O_4}$solution and we have to find the molarity of the solution.
We know that;
Normality is the no. of gram equivalents of solute present in $1L$ of solution.
Thus, $0.2N$ means $0.2$ gram equivalents are present in $1L$ of solution.
Gram equivalents = $\dfrac{{Mass}}{{Eq.mass}}$
Equivalent mass of ${H_2}S{O_4}$ = $\dfrac{{Mol.mass}}{{n - factor}}$
n-factor for ${H_2}S{O_4}$ = $2$
Mol. mass for ${H_2}S{O_4}$ = $98gmo{l^{ - 1}}$
Thus, Equivalent mass = $\dfrac{{98}}{2}$
= $49$
Now, putting the value of the equivalent mass in the gram equivalent formula. We can find the mass;
Mass = Gram equivalents $ \times $ Equivalent mass
= $0.2 \times 49$
= $9.8g$
Now, we can find molarity ( which is the number of moles of solute present in $1L$ of solution)
Number of moles of ${H_2}S{O_4}$ = $\dfrac{{mass}}{{mol.mass}}$
= $\dfrac{{9.8}}{{98}}$
= $0.1$ moles
Molarity= $\dfrac{{moles}}{{vol}}$
= $\dfrac{{0.1}}{1}$
= $0.1M$
Thus, the molarity was found to be $0.1M$

Note:
A simple formula is available that relates molarity with normality. We can also use that formula to find the molarity. The formula is;
Normality= n-factor $ \times $ molarity.
We know the n-factor and we are given with normality. We can easily find the molarity from this formula.