Question

T23.5g of sodium carbonate, $N{{a}_{2}}C{{O}_{3}}$ is dissolved in enough water to make 250ml of solution. If sodium carbonate dissociate completely, the molar concentration of sodium ion $N{{a}^{+}}$ and carbonate ion $C{{O}_{3}}^{2-}$ are respectively: (Molar mass of $N{{a}_{2}}C{{O}_{3}}=106mo{{l}^{-1}}$)
(A) 0.955M and 1.910M
(B) 1.910M and 0.955M
(C) 1.90M and 1.910M
(D) 0.477M and 0.477M

Answer 1

Hint: The measure of the concentration of a chemical species of particular solute in a solution, in terms of the amount of substance per unit volume of solution is known as Molar concentration or Molarity or amount concentration or substance concentration.
$Molarity(M)=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution(in litres)}}$

Complete step by step solution:
-We will start the solution by calculating the molarity of the solute,
Molar mass of $N{{a}_{2}}C{{O}_{3}}=106mo{{l}^{-1}}$
Calculating the number of moles-
Given mass = 25.3 g
$\text{Number of moles}=\dfrac{\text{Given mass}}{\text{Molar mass}}=\dfrac{23.5}{106}=0.238mol$
From the formula of Molarity,
Volume of solution = 250mL = 0.25L
$Molarity(M)=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution(in litres)}}=\dfrac{0.238}{0.25}=0.955M$
-Next step is finding the number of ions-
The dissociation of Sodium carbonates takes place in the following way-
$N{{a}_{2}}C{{O}_{3}}\to 2N{{a}^{+}}+C{{O}_{3}}^{2-}$
It is clear from the above equation that, dissociation of one mole of $N{{a}_{2}}C{{O}_{3}}$gives 2 moles of$N{{a}^{+}}$and 1 mole of $C{{O}_{3}}^{2-}$.
-The last step is the calculation of the concentration of sodium and carbonate ion-
Concentration of $N{{a}^{+}}$ ions$=2\times 0.955M=1.91M$
Concentration of $C{{O}_{3}}^{2-}$ ions$=1\times 0.955M=0.955M$

So, the correct answer is option (B).

Note: You should not get confused between molarity and molality.
(i) Molarity of a given solution is defined as the total number of moles of solute per litre of the solution, while molality is defined as the total moles of a solute contained in a kilogram of a solvent.
(ii) Mathematical expression for molarity is –
$Molarity(M)=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution(in litres)}}$
While the mathematical expression of molality is-
$Molality(m)=\dfrac{\text{Number of moles of solute}}{\text{Mass of solvent (in kg)}}$
(iii) Molarity depends on the volume of the whole solution while molality depends on the mass of the solvent.
(iv) Molarity has the unit as moles/litre while molality has the unit as moles/kg.
However, molarity and molality have a relationship which is given as-
$m=\dfrac{M\times 1000}{(d\times 1000)-M({{M}^{'}})}$
where’ is the molality
‘M’ is the molarity
‘d’ is the density
‘${{M}^{'}}$’ is the molar mass of the solute.