Answer
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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.
\[18g\to 6.022\times {{10}^{23}}\]
\[1g\to \dfrac{6.022\times {{10}^{23}}}{18}\]
\[0.05g\to \dfrac{6.022\times {{10}^{23}}}{18}\times 0.05=1.67\times {{10}^{21}}\]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 \[\dfrac{1.67\times {{10}^{21}}}{3600}\]=\[46.38\times {{10}^{16}}\] 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.
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.
\[18g\to 6.022\times {{10}^{23}}\]
\[1g\to \dfrac{6.022\times {{10}^{23}}}{18}\]
\[0.05g\to \dfrac{6.022\times {{10}^{23}}}{18}\times 0.05=1.67\times {{10}^{21}}\]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 \[\dfrac{1.67\times {{10}^{21}}}{3600}\]=\[46.38\times {{10}^{16}}\] 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.
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