Question

Calculate the percentage of free volume available in 1 mole of gaseous water at 1 atm and $100^\circ C$. The density of liquid ${H_2}O$ at $100^\circ C$ is 0.958g/mL.

Answer 1

Hint:The volume of the solution can be calculated by using the ideal gas equation. The behavior of gas is described by the ideal gas law which gives the relation between pressure, volume, number of moles and temperature.

Complete step by step answer:
Given,
Number of moles is 1 mole.
Pressure is 1 atm.
Temperature is $100^\circ C$or 373 K.
Density is 0.958g/mL.
For a gaseous water, the ideal gas equation is given by the formula shown below.
$PV = nRT$
Where,
P is the pressure.
V is the volume.
n is the number of moles.
R is the universal gas constant.
T is the temperature.
Substitute the value in the equation.
$V = \dfrac{{nRT}}{P}$
$\Rightarrow V = \dfrac{{1 \times 0.0821 \times 373}}{1}$
$\Rightarrow V = 30.62L$
The volume is calculated by the formula.
$D = \dfrac{M}{V}$
Where,
D is the density.
M is the mass.
V is the volume.
The volume of 1 mole liquid water $= \dfrac{{18}}{{0.958}}$
$\Rightarrow$ The volume of 1 mole liquid water $= 18.789$mL or $18.79 \times {10^{ - 3}}$L
The volume percentage is calculated by the formula.
$V\% = \dfrac{{{V_s}}}{{{V_T}}} \times 100$
Where,
${V_S}$ is volume of solute.
${V_T}$ is the volume of solution.
To calculate the volume percentage, substitute the value in the equation.
$V\% = \dfrac{{18.79 \times {{10}^{ - 3}}}}{{30.62}} \times 100$
$V\% = 0.0614\%$
The percentage free volume $= 100 - 0.0614$
$\Rightarrow$The percentage free volume $= 99.938\%$.
Therefore, the percentage of free volume available in 1 mole of gaseous water at 1 atm and $100^\circ C$ is 99.938%.

Note:
The volume percentage is defined as the measure of the substance or solute concentration not the solvent concentration in the total volume of solution and in ideal gas law, the volume denotes the total volume of gaseous liquid.