Question

If the concentration of water vapour in the air is 1% and the total atmospheric pressure equals 1 atm, then the partial pressure of water vapour is?
(A) 0.1 atm
(B) 1 mm Hg
(C) 7.6 mm Hg
(D) 100 atm

Answer 1

Hint: Approach this question with the laws and relationship provided by scientist John Dalton about the partial pressure of gases. Dalton's law which is also called Dalton's law of partial pressures, where he stated that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of each gas in the mixture.
Formula used:
We will use the following relationship in this solution:-
${{p}_{i}}={{p}_{total}}\times {{x}_{i}}$
where,
${{p}_{i}}$= partial pressure of a gas
${{p}_{total}}$= Total pressure of the mixture of non-reacting gas
${{x}_{i}}$= mole dfraction of a gas

Complete answer:
Let us first study about Dalton’s Law partial pressure:-
-This law states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases where partial pressure is the pressure exerted by a specific gas in a mixture. The mathematical relationship given is as follows:-
-${{p}_{total}}={{p}_{1}}+{{p}_{2}}+{{p}_{3}}+{{p}_{4}}+p........\text{so on}$
-${{p}_{i}}={{p}_{total}}\times {{x}_{i}}$
where,
${{p}_{i}}$= partial pressure of a gas
${{p}_{total}}$= Total pressure of the mixture of non-reacting gas
${{x}_{i}}$= mole dfraction of a gas
The values provided are as follows:-
${{p}_{total}}$= 1 atm
Moles of water vapour = 1 mole
Total moles in a mixture of air = 100 moles
${{x}_{{{H}_{2}}O}}=\dfrac{1mole}{100moles}=0.01$
Putting all the values in ${{p}_{i}}={{p}_{total}}\times {{x}_{i}}$, we get:-
$\begin{align}
  & \Rightarrow {{p}_{{{H}_{2}}O}}={{p}_{total}}\times {{x}_{{{H}_{2}}O}} \\
 & \Rightarrow {{p}_{{{H}_{2}}O}}=1atm\times 0.01 \\
 & \Rightarrow {{p}_{{{H}_{2}}O}}=0.01atm \\
\end{align}$
-As we can see that few options are given in different units of pressure, so we must calculate the answer in that unit as well.
1atm = 760mm Hg
So $0.01atm=0.01atm\times \dfrac{760mmHg}{1atm}=7.6mmHg$

Hence, the partial pressure of water vapour is: (C) 7.6 mm Hg

Note:
-While solving such questions, always remember to follow the methods of unit conversion along with the calculation as per the requirement.
-Also read the question carefully and look for the units in which the answer is asked for.
-Regarding Dalton’s law, it was assumed that gas molecules took up no volume and also molecules had no intermolecular interaction.