Question

Find the molality of H2SO4 solution whose specific gravity is 1.98 gm/ml and 90% by volume
H2SO4.

Answer 1

Answer on Question #41589 - Chemistry - Other

Question

Find the molality of $\mathrm{H}_2\mathrm{SO}_4$ solution whose specific gravity is $1.98~\mathrm{g / ml}$ and $90\%$ by volume $\mathrm{H}_2\mathrm{SO}_4$.

Solution

Molality (b) of a solution is defined as the amount of substance (in mol) of solute ($n_{\text{solute}}$) divided by the mass (in kg) of the solvent ($m_{\text{solvent}}$):

$$b = \frac {n _ {\text {solute}}}{m _ {\text {solvent}}}$$

Let us assume we have $1000\mathrm{ml}$ of the solution. Mass of the solution:

$$m _ {\text {solution}} = V _ {\text {solution}} \cdot \rho_ {\text {solution}} = 1000 \mathrm{ml} \cdot 1.98 \mathrm{g} / \mathrm{ml} = 1980 \mathrm{g}$$

Volume of solvent:

$$V _ {\text {solvent}} = \frac {V _ {\text {solution}} \cdot (100 - 90)}{100} = 1000 \mathrm{ml} \cdot 0.1 = 100 \mathrm{ml}$$

Density of the solvent (water) is $1.00\mathrm{g / ml}$, so the mass of the solvent:

$$m _ {\text {solvent}} = V _ {\text {solvent}} \cdot \rho_ {\text {solvent}} = 100 \mathrm{ml} \cdot \frac {1.00 \mathrm{g}}{\mathrm{ml}} = 100 \mathrm{g} = 0.1 \mathrm{kg}$$

Mass of the solute:

$$m _ {\text {solute}} = m _ {\text {solution}} - m _ {\text {solvent}} = 1980 \mathrm{g} - 100 \mathrm{g} = 1880 \mathrm{g}$$

Number of moles of the solute ($\mathsf{M}_{\text{solute}} = \mathsf{M}_{\text{H}_2\text{SO}_4} = 98~\mathrm{g/mol}$):

$$n _ {\text {solute}} = \frac {m _ {\text {solute}}}{M _ {\text {solute}}} = \frac {1880 \mathrm{g}}{98 \mathrm{g/mol}} = 19.18 \mathrm{mol}$$

Molality of the solution:

$$b = \frac {n _ {\text {solute}}}{m _ {\text {solvent}}} = \frac {19.18 \mathrm{mol}}{0.1 \mathrm{kg}} = 191.8 \mathrm{mol} / \mathrm{kg}$$

Answer: 1918 mol/kg

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