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

(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 H_{2}SO_{4} and has a mass of 1.046 kg:

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

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

Mass of 1.00 L soln = 1.00 L × 1.046 \frac{kg}{L} = 1.046 kg

Subtracting the mass of H_{2}SO_{4} 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 H_{2}O in 1.00 L soln = (1.046 kg) – (0.0736 kg) = 0.972 kg H_{2}O

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

Thus, the mole fraction of H_{2}SO_{4} is

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} = 0.0137

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

Mass % of H_{2}SO_{4}  =  \frac{0.0736  kg  H_{2}SO_{4}}{1.046  kg  total} × 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 H_{2}SO_{4} dissolved in it, 1.00 kg of water would have 0.772 mol of H_{2}SO_{4} dissolved in it:

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

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