Question

250 mL of a sodium carbonate solution contains 2.65 g of $N{a}_{2}C{O}_3$. If 10 mL of this solution is diluted to 1 L, what is the concentration of the resultant solution?
(Molar mass of $N{a}_{2}C{O}_3$ = 106 g/mol)

• A. 0.001 M
• B. 0.01 M
• C. 0.1 M
• D. ${10}^{-4}$ M

Answer 1

The correct option is A 0.001 M

$\text{Number of moles}=\frac{\text{given mass}}{\text{molar mass}}=\frac{2.65}{106}=0.025mol$

$\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution in mL}}\times 1000$

$\text{Molarity}=\frac{0.025}{250}\times 1000=0.1M$
Conserving moles before and after dilution:
$M_1{V}_1=M_2{V}_2$
Where $M_2$ and $V_2$ are the final concentration and volume, respectively.
So, $(0.1\times 10)=M_2\times 1000$
$M_2=0.001M$
Therefore, the final concentration of the solution is $0.001M$.