Question

250 mL of a sodium carbonate solution contains 2.65 g of $N{a}_{2}C{O}_3.$ If this solution is diluted to 1 L, what is the molarity of the resultant solution?

• A. 0.025 M
• B. 0.250 M
• C. 0.055 M
• D. 0.067 M

Answer 1

The correct option is A 0.025 M

$\text{Moles of}N{a}_{2}C{O}_3=\frac{\text{given mass}}{\text{molar mass}}=\frac{2.65}{106}=0.025$

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

$\text{Molarity}=\frac{0.025}{250}\times 1000=0.1M$

250 ml of 0.1 M is diluted to 1 L
$M_1{V}_1=M_f{V}_f$
$\Rightarrow (0.1\times 250)=(M\times 1000)$
$\Rightarrow M=0.025$ M