# 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.