During the day, 50 g of water evaporated how many molecules each second left the surface of the water

Problem data: m (mass of evaporated water) = 50 g (in SI m = 0.05 kg); t (elapsed time) = 1 day. (in SI system t = 86400 s).

Reference values: Na (Avogadro’s number) = 6.02 * 10 ^ 23 mol-1; M (molar mass of water) = 0.018 kg / mol.

The number of evaporated molecules in 1 s is determined by the formula: n1 = N / t = m * Na / (t * M).

Let’s calculate: n1 = N / t = 0.05 * 6.02 * 10 ^ 23 / (86400 * 0.0 ^ 18) = 19.35 * 10 ^ 18 molecules.

Answer: Every second, 19.35 * 10 ^ 18 molecules left the surface of the water.