Question

$250{\text{ mL}}$ of sodium carbonate solution contains $2.65{\text{ g}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. If $10{\text{ mL}}$ of this solution is diluted to $500{\text{ mL}}$, the concentration of the diluted solution will be?
A) $0.01{\text{ M}}$
B) $0.001{\text{ M}}$
C) $0.05{\text{ M}}$
D) $0.002{\text{ M}}$

Answer 1

Hint: To solve this we must know the molar mass of sodium carbonate. The molar mass of sodium carbonate i.e. ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is $106{\text{ g/mol}}$. First calculate the number of moles in $2.65{\text{ g}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. Then calculate the molarity of the solution. Then calculate the moles in $10{\text{ mL}}$ solution. Then using the moles calculate the molarity when the solution is diluted to $500{\text{ mL}}$.

Formulae Used:
1.${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
2. ${\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Volume of solution}}\left( {\text{L}} \right)}}$

Complete step-by-step answer:
We are given that $250{\text{ mL}}$ of sodium carbonate solution contains $2.65{\text{ g}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$.
We know that the number of moles is the ratio of mass to the molar mass. Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
The molar mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is $106{\text{ g/mol}}$. Thus,
${\text{Number of moles of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = \dfrac{{{\text{2}}{\text{.65 g}}}}{{106{\text{ g/mol}}}} = 0.025{\text{ mol}}$
Thus, $2.65{\text{ g}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ contains $0.025{\text{ mol}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$.
We know that molarity of a solution is the number of moles of solute per litre of solution. Thus,
${\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Volume of solution}}\left( {\text{L}} \right)}}$ …… (1)
$250{\text{ mL}} = 250 \times {10^{ - 3}}{\text{ L}}$ of sodium carbonate solution contains $0.025{\text{ mol}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. Thus,
${\text{Molarity}} = \dfrac{{0.025{\text{ mol}}}}{{250 \times {{10}^{ - 3}}{\text{ L}}}} = 0.1{\text{ M}}$
Thus, the molarity of $250{\text{ mL}}$ of sodium carbonate solution that contains $2.65{\text{ g}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$is $0.1{\text{ M}}$.
Now, calculate the number of moles of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ in $10{\text{ mL}}$ of $0.1{\text{ M}}$ solution as follows:
Rearrange equation (1) for the number of moles. Thus,
${\text{Number of moles of solute}}\left( {{\text{mol}}} \right) = {\text{Molarity}}\left( {\text{M}} \right) \times {\text{Volume of solution}}\left( {\text{L}} \right)$
We have $10{\text{ mL}} = 10 \times {10^{ - 3}}{\text{ L}}$ of $0.1{\text{ M}}$ solution. Thus,
${\text{Number of moles of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = 0.1{\text{ M}} \times 10 \times {10^{ - 3}}{\text{ L}} = 0.001{\text{ mol}}$
Thus, $10{\text{ mL}}$ solution of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ contains $0.001{\text{ mol}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$.
Now, the $10{\text{ mL}}$ solution is diluted to $500{\text{ mL}}$. Calculate the molarity of the solution using equation (1) as follows:
$500{\text{ mL}} = 500 \times {10^{ - 3}}{\text{ L}}$ of sodium carbonate solution contains $0.001{\text{ mol}}$ of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. Thus,
${\text{Molarity}} = \dfrac{{0.001{\text{ mol}}}}{{500 \times {{10}^{ - 3}}{\text{ L}}}} = 0.002{\text{ M}}$
Thus, the concentration of the diluted solution is $0.002{\text{ M}}$.

Thus, the correct option is (D) $0.002{\text{ M}}$.

Note: Molarity is used to express the concentration of solution. Remember that molarity is the number of moles of solute per litre of solution i.e. the amount of substance in a certain volume of solution. The unit of molarity is ${\text{mol }}{{\text{L}}^{ - 1}}$ or ${\text{M}}$. Molarity is also known as the molar concentration of a solution.