# $25.3g$ of sodium carbonate $N{a_2}C{O_3}$ is dissolved in enough water to make $250mL$ of solution. If sodium carbonate dissociates completely molar concentration of sodium ions and carbonate ions are respectively (molar mass of $N{a_2}C{O_3}$ is $106gmo{l^{ - 1}}$)

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**Hint:**In the question they gave volume, mass of the solution. By that, you can calculate the molarity of the solution. Molarity is defined as the number of moles of solute per liter of solution.

Molarity of solution $ = $ moles of the solute$/$litres of solution

**Complete answer:**Mass of the solute $ = $$25.3g$

Molar mass of solution $ = $$106gmo{l^{ - 1}}$

Volume of solution $ = $$250mL$

If the solute is dissolved in water then the solute splits into ions. Like that the sodium carbonate dissociates completely then splits into sodium and carbonate ions.

$N{a_2}C{O_3} \to 2N{a^ + } + C{O_3}^ - $

Molar concentration is nothing but molarity of solution

Molarity of solution $ = $ moles of the solute$/$litres of solution

$M = n/v$

Number of moles of solute$ = $$25.3/106 = 0.239moles$

Molarity of solution $ = $$0.239/0.25 = 0.956M$

As sodium carbonate dissociates into two sodium ions, the total molar concentration is calculated for two sodium ions.

Concentration of $N{a^ + } = $ $2 \times 0.956 = 1.912M$

Concentration of $C{O_3}^ - = $$0.956M$

**Note:**

Concentrate more on balancing the equations and be clear about the molarity, molality, and normality of solutions. Molarity is different from molality so don’t get confused about the terms. The molarity of a solution is related to its molecular weight of the solution and the molality of the solution is related to its equivalent weight of the solution.