Question

What is the mass of a water molecule in gram? How many molecules are present in one drop of pure water which weighs 0.05 g ? If the same drop of water evaporates in one hour, calculate the number of molecules leaving the liquid surface per second.

Answer 1

Hint: Water molecule contains two hydrogen atoms and one oxygen atom. And we know that there are $$6.022\times {10}^{23}$$ molecules in one mole water.

Complete step-by-step answer:
In a water molecule there are three atoms, one oxygen atom and two hydrogen atoms.
So, the molecular mass of one water molecule = 2 x (molar mass of one hydrogen atom) + 1 x (molar mass of oxygen atom).
And we know that molar mass of hydrogen = 1 g/mol,
Molar mass of oxygen atom = 16 g/mol.
Putting these values of molar mass in above equation for calculating molecular mass of water:
molecular mass of one water molecule = $$2\times (1)+16$$
                         = 18 g/mol
Mass of the water molecule in gram = 18/ $$6.022\times {10}^{23}$$
                            =$$2.988\times {10}^{-23}g$$
Now, we know that in 18 g of water there are $$6.022\times {10}^{23}$$ molecules.
$$18g\to 6.022\times {10}^{23}$$
$$1g\to {\displaystyle \frac{6.022\times {10}^{23}}{18}}$$
$$0.05g\to {\displaystyle \frac{6.022\times {10}^{23}}{18}}\times 0.05=1.67\times {10}^{21}$$molecules
It is given that evaporation of 0.05 g water droplets takes 1 hour.
$$1.67\times {10}^{21}(0.05g)\to 1hour$$
Or in 3600 seconds $$1.67\times {10}^{21}$$ are evaporating or leaving the surface. So, in one second $${\displaystyle \frac{1.67\times {10}^{21}}{3600}}$$=$$46.38\times {10}^{16}$$ molecules are evaporating or leaving the surface.
So, the answer is $$46.38\times {10}^{16}$$ molecules.

Note: Another method of calculating evaporation per seconds is that initially we can calculate the number evaporation rate in grams and we can convert it into a number of molecules.