Solution concentration — molarity, molality, normality, mole fraction conversion

Question

A solution contains 49 g of H $_2$ SO $_4$ in 500 mL of solution. The density of the solution is 1.2 g/mL. Calculate the molarity, molality, and mole fraction of H $_2$ SO $_4$ .

(JEE Main / CBSE 12 — Solutions)

Concentration Unit Conversion

flowchart TD
    A["Concentration Problem"] --> B{What is given?}
    B -->|Moles + Volume of solution| C["Molarity (M) = n/V in L"]
    B -->|Moles + Mass of solvent| D["Molality (m) = n/kg solvent"]
    B -->|Equivalents + Volume| E["Normality (N) = eq/V"]
    B -->|Moles of each component| F["Mole fraction = n_i / n_total"]
    C -->|Need molality?| G["Find mass of solvent from density"]
    D -->|Need molarity?| H["Find volume from density"]
    C --> I["N = n-factor x M"]

Solution — Step by Step

Molar mass of H $_2$ SO $_4$ = 2(1) + 32 + 4(16) = 98 g/mol

Moles of H $_2$ SO $_4$ = $\dfrac{49}{98} = 0.5$ mol

Volume of solution = 500 mL = 0.5 L

M = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.5}{0.5} = \mathbf{1 \text{ M}}

We need the mass of solvent (water), not the solution.

Mass of solution = Volume $\times$ Density = 500 $\times$ 1.2 = 600 g

Mass of solvent = Mass of solution - Mass of solute = 600 - 49 = 551 g = 0.551 kg

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.5}{0.551} = \mathbf{0.907 \text{ mol/kg}}

Moles of H $_2$ SO $_4$ = 0.5 mol

Moles of H $_2$ O = $\dfrac{551}{18} = 30.61$ mol

\chi_{\text{H}_2\text{SO}_4} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} = \frac{0.5}{0.5 + 30.61} = \frac{0.5}{31.11}

\boxed{\chi_{\text{H}_2\text{SO}_4} = 0.0161}

Why This Works

Different concentration units are useful in different contexts:

Molarity depends on volume (which changes with temperature) — used in volumetric analysis
Molality depends on mass of solvent (temperature-independent) — used in colligative property calculations
Mole fraction is dimensionless — used in Raoult’s law and gas calculations

The key conversion step is always: find the mass of solvent separately from the mass of solution. Density connects mass and volume.

Alternative Method — Quick Conversion Formulas

If you know molarity $M$ and density $d$ (in g/mL), molar mass of solute $M_B$ :

m = \frac{1000 \times M}{1000d - M \times M_B}

For our problem: $m = \dfrac{1000 \times 1}{1000(1.2) - 1(98)} = \dfrac{1000}{1102} = 0.907$ mol/kg

For JEE and NEET, the interconversion between molarity and molality is a guaranteed numerical. The shortcut formula above saves 2-3 minutes compared to the step-by-step method. But always know the step-by-step for CBSE board exams, where working matters more than speed.

Common Mistake

The classic error: using mass of solution instead of mass of solvent for molality. Molality uses mass of solvent (water), while molarity uses volume of solution. Another frequent mistake: forgetting to subtract the mass of solute from the mass of solution to get the mass of solvent. If density is 1.2 g/mL and volume is 500 mL, the solution mass is 600 g — but the solvent mass is 600 - 49 = 551 g, not 600 g.