What amount of the substance is contained in 50 g of a 20% solution of drinking salt? (NaHCO3)

Given:
m solution (NaHCO3) = 50 g
ω (NaHCO3) = 20%

Find:
n (NaHCO3) -?

1) m (NaHCO3) = ω (NaHCO3) * m solution (NaHCO3) / 100% = 20% * 50/100% = 10 g;
2) M (NaHCO3) = Mr (NaHCO3) = Ar (Na) * N (Na) + Ar (H) * N (H) + Ar (C) * N (C) + Ar (O) * N (O ) = 23 * 1 + 1 * 1 + 12 * 1 + 16 * 3 = 84 g / mol;
2) n (NaHCO3) = m (NaHCO3) / M (NaHCO3) = 10/84 = 0.1 mol.

Answer: The amount of NaHCO3 substance is 0.1 mol.