Question

The solute having lowest molar concentration is?
(A) $ 1.0N{\text{ HCl}} $
(B) $ {\text{0}}{\text{.4N }}{{\text{H}}_2}S{O_4} $
(C) $ 0.1N{a_2}C{O_3} $
(D) $ 1N{\text{ NaOH}} $

Answer 1

Hint :Molar concentration is the measure of the amount of a chemical species in a solution, generally, it measures amount of solute present in the solution, in terms of amount of substance per unit volume of a solution.

Complete Step By Step Answer:
molarity and normality can be interrelated, a general formula
Normality is equivalent factor times of the molarity of the solution.
Normality is, $ N = Molarity \times equivalent{\text{ factor}} $
 $ N = M \times {n_{eq}} $
Now, this formula will help us to calculate the molar concentration of each given normal concentration.
Since, we can derive the above formula as –
 $ Molarity = \dfrac{{Normality}}{{Equivalent{\text{ factor}}}} $
Here, Equivalent factors ( $ {n_{eq}} $ ) are the number of ionisable atoms or the one which will precipitate.
So, looking at our first option i.e. $ 1.0N{\text{ HCl}} $ ,
 $ Molarity = \dfrac{{Normality}}{{Equivalent{\text{ factor}}}} $
 $ Molarity = \dfrac{1}{1} = 1M $
Since, here ionisable atom hydrogen is $ {\text{1}} $ .
So, molarity of $ 1.0N{\text{ HCl}} $ is $ {\text{1M}} $
Similarly For, $ {\text{0}}{\text{.4N }}{{\text{H}}_2}S{O_4} $
 $ Molarity = \dfrac{{0.4}}{2} = 0.2M $
Here ionisable atom hydrogen is $ {\text{2}} $
So, molarity of $ {\text{0}}{\text{.4N }}{{\text{H}}_2}S{O_4} $ is $ {\text{0}}{\text{.2M}} $
Coning the next option, $ 0.1N{a_2}C{O_3} $
 $ Molarity = \dfrac{{0.1}}{2} = 0.05M $
So, molarity of $ 0.1N{a_2}C{O_3} $ is $ {\text{0}}{\text{.05M}} $
Our last option,
 $ 1N{\text{ NaOH}} $
 $ Molarity = \dfrac{1}{1} = 1M $
So, molarity of $ 1N{\text{ NaOH}} $ is $ {\text{1M}} $
Thus, after solving all the equations, we can conclude our result.
So, $ 0.1N{a_2}C{O_3} $ has the lowest molar concentration among the given options.
Therefore, option C is the correct answer.

Note :
Normality can be defined as the mole or gran number of equivalents of the solute present in the one litre of the solution. The gram equivalents are the ionisable group or the species which are available for the reaction in the solution. The SI unit of Normality is $ {\text{mol }}{{\text{m}}^{ - 3}} $ .