Question

The solution having lowest molar concentration is:
(A) $ 1.0N $ $ HCl $
(B) $ 0.4N $ $ {H_2}S{O_4} $
(C) $ 0.1N $ $ N{a_2}C{O_3} $
(D) $ 1N $ $ NaOH $

Answer 1

Hint: Molar concentration is also known as molarity of solution which is used frequently to express the concentration of solution. Molarity and normality are two independent properties but have a mathematical relation with each other.

Complete Step By Step Answer:
Normality is used to express the concentration of any solution represented as $ \left( {\rm N} \right) $ . Normality is defined as a ratio of gram equivalent of initial solute taken $ {N_{\left( {eq} \right)}} $ to volume of solution considered in litres $ {V_{\left( l \right)}} $ .
 $ \left( {\rm N} \right) $ $ = \dfrac{{{N_{\left( {eq} \right)}}}}{{{V_{\left( l \right)}}}} $ .
Gram equivalent mass of compound is calculated by dividing mass of solute by equivalent mass of solute.
 $ {N_{\left( {eq} \right)}} = \dfrac{m}{{Eq}} $
Equivalent weight of any acid or base is calculated by taking the ratio of molecular weight of acid to acidity of the acid.
 $ Eq = \dfrac{M}{n} $
Where $ Eq $ ,is equivalent weight of the compound
 $ M $ , is molecular weight of compound
 $ n $ , is basicity of acid or acidity of base or valency factor
Now we modify the formula of normality in terms of molarity
 $ \left( {\rm N} \right) $ $ = \dfrac{{m \times n}}{{M \times V}} $ .
Where, $ \dfrac{m}{{M \times V}} $ is the ratio of mass to molecular mass which is also known as molarity of solution.
So, the final relation between normality and molarity is expressed as:
 $ {\rm N} = M \times n $
 $ \to $ For $ 0.1{\rm N} $ $ HCl $ , as we know hydrochloric acid is monobasic acid because it has only one hydrogen atom to release. So, the value of valency factor of $ HCl $ is one.
 $ 0.1 = M \times 1 $
After solving the above equation, we get
 $ M = \dfrac{{0.1}}{1} $
Therefore molarity of $ 0.1{\rm N} $ $ HCl $ is $ 0.1M $
 $ \to $ For $ 0.4{\rm N} $ $ {H_2}S{O_4} $ , as we know sulphuric acid is dibasic acid because it has two hydrogen atoms to release. So, the value of the valency factor of $ {H_2}S{O_4} $ is two.
 $ 0.4 = M \times 2 $
After solving the above equation, we get
 $ M = \dfrac{{0.4}}{2} $
Therefore molarity of $ 0.1{\rm N} $ $ HCl $ is $ 0.2M $
 $ \to $ For $ 0.1{\rm N} $ $ N{a_2}C{O_3} $ , as we know sodium carbonate is a diacidic base because it has two carbonate atoms to release. So, the value of the valency factor of $ N{a_2}C{O_3} $ is two.
 $ 0.1 = M \times 2 $
After solving the above equation, we get
 $ M = \dfrac{{0.1}}{2} $
Therefore, the molarity of $ 0.1{\rm N} $ $ N{a_2}C{O_3} $ is $ 0.05M $ .
 $ \to $ For $ 1.0{\rm N} $ $ NaOH $ , as we know sodium hydroxide is monoacidic base because it has only one hydroxide molecule to release. So, the value of the valency factor of $ NaOH $ is one.
 $ 1.0 = M \times 1 $
After solving the above equation, we get
 $ M = \dfrac{{1.0}}{1} $
Therefore, the molarity of $ 1.0{\rm N} $ $ NaOH $ is $ 1.0M $ .
 $ \Rightarrow $ After the above discussion we conclude that $ 0.1{\rm N} $ $ N{a_2}C{O_3} $ has the lowest value of molar concentration of $ 0.05M $ .
Therefore, option $ \left( C \right) $ is the correct option.

Note:
Value of the valency factor for salt is calculated by the calculation of total positive valency of combining atoms, and another approach to calculate the valency factor for salt is by determining the total number of electrons which are either lost or gained by the salt.