Question 39:(I) How many moles of water are there in 1.000 L at STP? How many molecules?

The ideal gas law gives the relation between the pressure (P), volume (V), and temperature (T) of n moles of the ideal gas by the following expression:

\(PV = nRT\)

Here, R is the universal gas constant.

In terms of the number of molecules (N) of the gas, the ideal gas law is written as

\(PV = \frac{N}{{{N_{\rm{A}}}}}RT\).

Here, \({N_{\rm{A}}}\)is the number of molecules in one mole of the substance and is known as Avogadro’s number. Its value is\(6.02 \times {10^{23}}\).

The volume of water is \(V = 1.00\;{\rm{L}} = 1.00 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{3}}}\)

The density of water is \(\rho = 1.00 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^3}\)

The mass of the given amount of water is\(M = \rho \times V\).

Substitute the values in the above equation.

\(\begin{array}{c}M = \left( {1.00 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3}} \right) \times \left( {1.00 \times {{10}^{ - 3}}\;{{\rm{m}}^{\rm{3}}}} \right)\\ = 1.00\;{\rm{kg}}\end{array}\)

One molecule of water contains two atoms of oxygen and one atom of hydrogen. Thus, the molecular mass of one water molecule is

\(\begin{array}{c}{M_0} = \left( {2 \times 1} \right) + \left( {16} \right)\\ = 18\;{\rm{g}}\\ = 18 \times {10^{ - 3}}\;{\rm{kg}}{\rm{.}}\end{array}\)

The number of moles of water is equal to itsmass divided by its molecular mass, i.e.,

\(\begin{array}{c}n = \frac{M}{{{M_0}}}\\ = \frac{{1.00\;{\rm{kg}}}}{{18 \times {{10}^{ - 3}}\;{\rm{kg}}}}\\ = 55.55.\end{array}\)

Thus, there are 55.55 moles in 1.00 L of water.

The number of molecules in 1.00 L of water is

\(\begin{array}{c}n = \frac{N}{{{N_{\rm{A}}}}}\\N = n{N_{\rm{A}}}.\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}N = 55.55 \times 6.02 \times {10^{23}}\;{\rm{molecules}}\\ = 334.41 \times {10^{23}}\;{\rm{molecules}}\\ = 3.34 \times {10^{25}}\;{\rm{molecules}}\end{array}\)

Thus, there are \(3.34 \times {10^{25}}\;{\rm{molecules}}\) in 1.00 L of water at STP.