Question

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

Short answer

In1.00 Lof water at STP, there are 55.55 moles and $3.34\times {10}^{25}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}$.

Step-by-step solution

Step 1:Ideal gas law in terms of 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

$PV=\frac{N}{N_{\mathrm{A}}}RT$.

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

Given information 

The volume of water is $V=1.00\mathrm{L}=1.00\times {10}^{-3}{\mathrm{m}}^3$

The density of water is $\rho =1.00\times {10}^3\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$

Determination of the mass of water 

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

Substitute the values in the above equation.

$\begin{array}{c}M=\left(1.00\times {10}^3\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\right)\times \left(1.00\times {10}^{-3}{\mathrm{m}}^3\right)\\ =1.00\mathrm{k}\mathrm{g}\end{array}$

Determination of the number of moles of water

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

$\begin{array}{c}{M}_0=\left(2\times 1\right)+\left(16\right)\\ =18\mathrm{g}\\ =18\times {10}^{-3}\mathrm{k}\mathrm{g}.\end{array}$

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

$\begin{array}{c}n=\frac{M}{M_0}\\ =\frac{1.00\mathrm{k}\mathrm{g}}{18\times {10}^{-3}\mathrm{k}\mathrm{g}}\\ =55.55.\end{array}$

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

Determination of the number of molecules of water 

The number of molecules in 1.00 L of water is

$\begin{array}{c}n=\frac{N}{N_{\mathrm{A}}}\\ N=n{N}_{\mathrm{A}}.\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}N=55.55\times 6.02\times {10}^{23}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\\ =334.41\times {10}^{23}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\\ =3.34\times {10}^{25}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\end{array}$

Thus, there are $3.34\times {10}^{25}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}$ in 1.00 L of water at STP.