Question

$25.3g$ of sodium carbonate is dissolved in enough water to make $250ml$ of solution .If sodium carbonate dissociates completely molar concentration of sodium ions $N{a^ + }$ and carbonate ions are respectively (molar mass of $N{a_2}C{O_3} = 106g/mol$ )
A.$0.955M{\text{ and 1}}{\text{.910M}}$
B.${\text{1}}{\text{.910M and }}0.955M$
C.$1.90M{\text{ and 1}}{\text{.910M}}$
D.$0.477M{\text{ and 0}}{\text{.477M}}$

Answer 1

Hint:
First we should be aware with the term moles and the way to find them so that we can work on them for the further steps. Now after getting the moles we have to get the molarity ( moles per unit volume/ litre ). For that we will use the moles we found earlier. After that we can give the balanced disintegration reaction, so that we can apply a unitary method for the further answer finding.

Complete step by step answer:
To find the number of moles in any of the compounds we have to divide the giver mass of the compound by the molar mass. To obtain the molar mass we have to add the molecular masses of the compound.
For the case of
$
  N{a_2}C{O_3} = \\
   = 2Na + C + 3O \\
   = 2(23) + 12 + 3(16) \\
   = 106g/mol \\
 $
Therefore the number of the moles would be :
$
   = \dfrac{{25.3}}{{106}} \\
   = 0.239mol \\
 $
Now, we have to find the molarity of the given sample for that we divide the moles of the compound with the volume given. Because molarity is mol per volume:
$
   = \dfrac{{0.239}}{{0.25}} \\
   = 0.956mol/litre \\
 $
Now we have to define the given reaction for the disintegration reaction :
$N{a_2}C{O_3} \to 2N{a^ + } + C{O_3}^{2 - }$
So, by the unitary method we get that every one mole of the sodium carbonate gives two moles of the Sodium cation.
Therefore the molarity would change accordingly.
Concentration of the given cation, $N{a^ + } = 2 \times 0.956 = 1.912M$
Therefore, the answer would be option B, ${\text{1}}{\text{.910M and }}0.955M$.

Note:
Molarity can be used to calculate the volume of solvent or the amount of solute. In this question, we have to use the unitary method. Although there are more methods for observing and finding molarity as per the required solution .