USING DENSITY TO CONVERT MOLARITY TO OTHER MEASURES OF CONCENTRATION A 0.750 M solution of H2SO4 in water has a density of 1.046 g/mL at 20 °C. What is the concentration of this solution in (a) mole fraction, (b) mass percent, and (c) molality? The molar mass of H2SO4 is 98.1 g/mol.

Question

USING DENSITY TO CONVERT MOLARITY TO OTHER MEASURES OF CONCENTRATION

A 0.750 M solution of [latex]H_{2}SO_{4}[/latex] in water has a density of 1.046 g/mL at 20 °C. What is the concentration of this solution in (a) mole fraction, (b) mass percent, and (c) molality? The molar mass of [latex]H_{2}SO_{4}[/latex] is 98.1 g/mol.

Accepted Answer

(a) Let’s pick an arbitrary amount of the solution that will make the calculations easy, say 1.00 L. Since the concentration of the solution is 0.750 mol/L and the density is 1.046 g/mL (or 1.046 kg/L), 1.00 L of the solution contains 0.750 mol (73.6 g) of [latex]H_{2}SO_{4}[/latex] and has a mass of 1.046 kg:

Moles of [latex]H_{2}SO_{4} [/latex] in 1.00 L soln = 0.750 [latex] \frac{mol}{L}[/latex] × 1.00 L = 0.750 mol

Moles of [latex] H_{2}SO_{4} [/latex] in 1.00 L soln [latex]= 0.750 mol × 98.1 \frac{g}{mol}[/latex] = 73.6 g

Mass of 1.00 L soln = 1.00 L × 1.046 [latex]\frac{kg}{L}[/latex] = 1.046 kg

Subtracting the mass of [latex]H_{2}SO_{4}[/latex] from the total mass of the solution gives 0.972 kg, or 54.0 mol, of water in 1.00 L of solution:

Mass of [latex]H_{2}O[/latex] in 1.00 L soln = (1.046 kg) - (0.0736 kg) = 0.972 kg [latex]H_{2}O[/latex]

Moles of [latex]H_{2}O[/latex] in 1.00 L of soln = 972 g × [latex]\frac{1 mol}{18.0 g} = 54.0 mol H_{2}O[/latex]

Thus, the mole fraction of [latex]H_{2}SO_{4}[/latex] is

[latex]X_{H_{2}SO_{4}} = \frac{0.750 mol H_{2}SO_{4}}{0.750 mol H_{2}SO_{4} + 54.0 mol H_{2}O}[/latex] = 0.0137

(b) The mass percent concentration can be determined from the calculations in part (a):

Mass % of [latex]H_{2}SO_{4} = \frac{0.0736 kg H_{2}SO_{4}}{1.046 kg total}[/latex] × 100% = 7.04%

(c) The molality of the solution can also be determined from the calculations in part (a). Since 0.972 kg of water has 0.750 mol of [latex]H_{2}SO_{4}[/latex] dissolved in it, 1.00 kg of water would have 0.772 mol of [latex]H_{2}SO_{4}[/latex] dissolved in it:

1.00 kg [latex]H_{2}O × \frac{0.750 mol H_{2}SO_{4}}{0.972 kg H_{2}O} = 0.772 mol H_{2}SO_{4}[/latex]

Thus, the molality of the sulfuric acid solution is 0.772 m.

Question 11.6: USING DENSITY TO CONVERT MOLARITY TO OTHER MEASURES OF CONCE...

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