The concentration of a solution may be defined as the amount of solute present in the given quantity of the solution.
The concentration of solution may be expressed in several ways as discussed below:
Mass Percentage
The mass percentage of a component in a given solution is the mass of the component per 100 g of the solution.
\[\text{Mass percentage of a component = }\frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} × 100\]
Molarity of a solution
The molarity of a solution is defined as the number of moles of the solute dissolved per litre of the solution.It is represented as ′M′. Mathematically,
\[Molarity = \frac{\text{Moles of solute}}{\text{Volume of the solution (in mL)}} × 1000\]
Molality of the solution
The molality of a solution is defined as the number of moles of the solute dissolved per 1000 g (or 1 kg) of the solvent.
A solution of urea (NH2CONH2) in water is labelled 3% (w/V). What is the molarity of this solution? (Molar mass of urea = 60 g mol−1)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.50 M
Mass by volume percentage (w/V) is the mass of solute dissolved in 100 mL of the solution. So a 3% (w/V) urea solution has 3 g of urea in every 100 mL of solution. Molarity needs the moles of solute per litre of solution.
\( \displaystyle n(\text{urea}) = \frac{3\ \mathrm{g}}{60\ \mathrm{g\ mol^{-1}}} = 0.050\ \mathrm{mol} \)
\( \displaystyle V = 100\ \mathrm{mL} = 0.100\ \mathrm{L} \)
\( \displaystyle \text{Molarity} = \frac{0.050\ \mathrm{mol}}{0.100\ \mathrm{L}} = 0.50\ \mathrm{mol\,L^{-1}} = 0.50\ \mathrm{M} \)
NCERT (2026-27 edition):
1.2 Expressing Concentration of Solutions – “(iii) Mass by volume percentage (w/V): Another unit which is commonly used in medicine and pharmacy is mass by volume percentage. It is the mass of solute dissolved in 100 mL of the solution.”
“(vi) Molarity: Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution,”
\( \displaystyle \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}} \qquad (1.8) \)