Question

The ratio of moles of $N{a}_{2}C{O}_3$ to total number of moles in a 100 mL aqueous solution of $N{a}_{2}C{O}_3$ is 1:5. If 90 g of water is present in 100 mL of solution, find the molarity of the solution.

• A. 12.5 M
• B. 1.2580 M
• C. 0.0125 M
• D. 0.1538 M

Answer 1

The correct option is A 12.5 M

Let the moles of sodium carbonate be denoted as ‘a’ mol
According to the given data:
90 g of water is present in 100 mL of the solution.
The given ratio of moles of sodium carbonate to the total number of moles in solution does not change.
Number of moles of water present in 100 mL of the solution $=\frac{90}{18}=5mol$
$\frac{molesofsodiumcarbonate}{totalnumberofmoles}=\frac{1}{5}$
$\frac{1}{5}=\frac{molesofsodiumcarbonate}{molesofsodiumcarbonate+molesofwater}$
⇒$\frac{1}{5}=\frac{a}{a+5}$
On solving, a = 1.25 mol
$molarity=\frac{molesofsolute}{volumeofsolutioninL}$
$molarity=\frac{1.250\times 1000}{100}=12.50M$