Determine the mass of an aluminum sulfate solution with a mass fraction of a solute of 15%
Determine the mass of an aluminum sulfate solution with a mass fraction of a solute of 15% required to obtain aluminum hydroxide weighing 2.34 g.
Find the amount of aluminum hydroxide substance by the formula:
M (Al (OH) 3) = 27 + 3 × 17 = 78 g / mol.
n = m: M.
n = 2.34 g: 78 g / mol = 0.03 mol.
Let’s find the quantitative ratios of substances.
Al2 (SO4) 3 = 2Al (OH) 3
For 1 mol of aluminum sulfate, there are 2 mol of aluminum hydroxide.
The substances are in quantitative ratios of 1: 2.
The amount of aluminum sulfate substance will be 2 times less than the amount of aluminum hydroxide substance.
1 / 2n (Al2 (SO4) 3) = n (2Al (OH) 3) = 0.03: 2 = 0.015 mol.
Let’s find the mass of aluminum sulfate by the formula:
m = n × M,
M (Al2 (SO4) 3) = 27 × 2 + 3 (32 + 64) = 342 g / mol.
m = 0.015 mol × 342 g / mol = 5.13 g.
Find the mass of the aluminum sulfate solution.
Mass fraction of a substance is calculated by the formula:
W = m (substance): m (solution) × 100%,
m (solution) = (m (substance): W) × 100%
m (solution) = (5.13 g: 15%) × 100% = 34.2 g.
Answer: m (solution) = 34.2 g.