To 500 ml of 32% HNO3 solution (p = 1.20 g / ml) 1 liter of water was added.

To 500 ml of 32% HNO3 solution (p = 1.20 g / ml) 1 liter of water was added. What is the mass fraction of HNO3 in the resulting solution?

1. Calculate the mass of the initial nitric acid solution:

mout (solution HNO3) = V (solution HNO3) * ρ (solution HNO3) = 500 * 1.2 = 600 g;

2. find the mass of the dissolved acid:

m (HNO3) = wout (HNO3) * mout (solution HNO3) = 0.32 * 600 = 192 g;

3.determine the mass of the solution obtained by dilution:

mpol (solution HNO3) = mout (solution HNO3) + madd (H2O) = 600 + 1000 = 1600 g;

4.Calculate the mass fraction of acid in the resulting solution:

wpol (HNO3) = m (HNO3): mpol (solution HNO3) = 192: 1600 = 0.12 or 12%.

Answer: 12%.



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