Question

The ratio of moles of A to total number of moles present in 150 mL of 20% (v/v) solution is 0.3. Density of the solvent is 0.12 g/mL and molar mass of the solvent is 14.4 g/mol. Calculate the molarity and molality of the solution respectively.

• A. 2.85 M, 29.76 m
• B. 29.76 m, 2.85 M
• C. 3.85 M, 39.76 m
• D. 39.76 M, 3.85 m

Answer 1

The correct option is A 2.85 M, 29.76 m

Let the number of moles of solute be $n_A$
20% (v/v) means 20 mL of solute is present in 100 mL of solution.
Given:
volume of the solution = 150 mL
Volume of solvent present in 150 mL solution $=150\times 0.8=120mL$
Since, $density=\frac{mass}{volume}$
Mass of solvent $=volume\times density=120mL\times 0.12=14.4g$
Number of moles of solvent in 150 mL $=\frac{14.4}{14.4}=1mol$
⇒$\frac{MoleofsoluteA}{Totalnumberofmoles}=\frac{n_A}{(n_A+n_{solvent})}=0.3$
⇒ $\frac{n_A}{(n_A+1)}=0.3$
On solving, $n_A=\frac{3}{7}$
$Molarityofsolution=\frac{molesofsolute}{volumeofsolutioninmL}\times 1000$
$Molarity=\frac{n_A}{volumeofsolutioninmL}=\frac{3}{7\times 150}\times 1000=2.85M$
$Molality=\frac{n_A}{\left(massofsolventinkg\right)}=\frac{3}{7\times 14.4}\times 1000=29.76m$