40 grams of water was added to 80 grams of a 3% solution. What is the mass fraction of the solute in the resulting solution.

1.m (substance) = m1 (solution) * w1 (substance) (in fractions);

m (substance) = 80 * 0.03 = 2.4 g;

2.after dilution with water, the mass of the initial solution increases:

m2 (solution) = m1 (solution) + m (water);

m2 (solution) = 80 + 40 = 120 g;

3.w2 (substance) = m (substance): m2 (solution);

w2 (substances) = 2.4: 120 = 0.02 or 2%.

Answer: 0.02 or 2%.