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

Hint: 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.