For 10 days, 100 g of water has completely evaporated. How many molecules, on average, escaped

For 10 days, 100 g of water has completely evaporated. How many molecules, on average, escaped from its surface in 1 second? The molar mass of water is 0.018 kg / mol.

1. Let’s translate the day into seconds.

10 days = 10 * 24 * 3600 = 864000 s.

2. Find the mass of water evaporated in one second and convert it into the amount of matter.

m1 (H2O) = mototal (H2O) / ttotal = 0.1 kg / 864000 s = 1.157 * 10 ^ -6 kg.

n1 (H2O) = m1 (H2O) / M (H2O) = 1.157 * 10 ^ -6 kg / 0.018 kg / mol = 6.43 * 10 ^ -5 mol.

3. Find the number of water molecules evaporated in one second.

N1 (H2O) = n1 (H2O) * NA = 6.43 * 10 ^ -5 mol * 6.02 * 10 ^ 23 mol-1 = 3.87 * 10 ^ 18.

Answer: N1 (H2O) = 3.87 * 10 ^ 18.