To a solution of sodium nitrate weighing 800 g and a mass fraction of 4% was added 20 g

To a solution of sodium nitrate weighing 800 g and a mass fraction of 4% was added 20 g of the same salt. The mass of salt in the resulting solution is?

The mass of sodium nitrate in the original solution:

m1 (NaNO3) = mr.-pa1 (NaNO3) * w1 (NaNO3) / 100 = 800 * 4/100 = 32 (g).

The mass of sodium nitrate in the resulting solution:

m2 (NaNO3) = m1 (NaNO3) + madd. (NaNO3) = 32 + 20 = 52 (g).

The mass of the resulting solution:

mr.-pa2 (NaNO3) = m.-pa1 (NaNO3) + madd. (NaNO3) = 800 + 20 = 820 (g).

Thus, the mass fraction of sodium nitrate in the resulting solution:

w2 (NaNO3) = (m2 (NaNO3) /mр.-pa2 (NaNO3)) * 100 = (52/820) * 100 = 6.34 = 6 (%).