Another 50 g of water was added to 450 g of a solution of nitric acid HNO3 with a mass fraction of 38%

Another 50 g of water was added to 450 g of a solution of nitric acid HNO3 with a mass fraction of 38%. What was the mass fraction of HNO3 in the solution?

1. Calculate the mass of nitric acid in the original solution:

mout (HNO3) = wout (HNO3) * mout (solution HNO3) = 0.38 * 450 = 171 g;

2. find the mass of the solution obtained by dilution:

mpol (solution HNO3) = mout (solution HNO3) + m (H2O) = 450 + 50 = 500 g;

3. Determine the mass fraction of acid in the resulting solution:

wpol (HNO3) = mout (HNO3): mpol (solution HNO3) = 171: 500 = 0.342 or 34.2%.

Answer: 34.2%.