Question

Density of a mixture of ${{O}_{2}}$ and ${{N}_{2}}$ at NTP is $1.3 g/L$. Calculate partial pressure of ${{O}_{2}}$.
A. 0.28 atm
B. O.38 atm
C. 0.48 atm
D. 0.18 atm

Answer 1

Hint: To calculate the partial pressure of a gas in a gas mixture when total pressure is given dalton’s law of partial pressure can be used. It states that partial pressure of a gas in a gaseous mixture will be equal to the product of its mole fraction and total pressure of the mixture. Total pressure of the mixture is given to us as 1 atm, so all we have to do is calculate mole fraction with the help of density and ideal gas equation.

Complete step by step solution:
NTP means normal temperature and pressure.
At NTP pressure is always equal to $1 atm$ and temperature is equal to $273.15K$
Ideal gas equation in terms of density:
$PM=dRT$
Here:
P is the pressure and given value of it is $1atm$
M is the molecular mass of gas mixture
d is the density of gas mixture whose given value is $1.3g/L$
R is universal gas constant whose value is $0.082 L-atm/K-mol$
T is the temperature whose given value is $273.15 K$
Put all these values into ideal gas equation
$1\times M=1.3\times 0.082\times 273$
$M=0.1066\times 273.15$
$M=29.12g$
Now let’s assume mole fraction of ${{O}_{2}}$in the mixture is ${{x}_{1}}$ and mole fraction of ${{N}_{2}}$ is ${{x}_{2}}$
${{x}_{1}}$+${{x}_{2}}$= 1
\[{{x}_{2}}=1-{{x}_{1}}\]
As we know
${{M}_{mix}}={{x}_{1}}{{M}_{{{o}_{2}}}}+{{x}_{2}}{{M}_{{{N}_{2}}}}$
$29.12={{x}_{1}}\times 32+(1-{{x}_{1}})28$
$29.12={{x}_{1}}\times 32+28-28{{x}_{1}}$
$29.12-28=4{{x}_{1}}$
$4{{x}_{1}}=1.12$
${{x}_{1}}=\dfrac{1.12}{4}$
${{x}_{1}}=0.28$
As total pressure of the system is 1 atm and we also know the mole fraction of oxygen, therefore we will used dalton’s law of partial pressure
${{P}_{{{o}_{2}}}}={{x}_{1}}{{P}_{total}}$
${{P}_{{{o}_{2}}}}=0.28\times 1$
${{P}_{{{O}_{2}}}}=0.28atm$
So partial pressure of oxygen is 0.28 atm

Correct answer for this question is Option (A).

Additional information:
Mole fraction of a gas in gaseous mixture is always equal to the ratio of its mole to the total number of gaseous moles present in the system.

Note: Oxygen is a colorless, tasteless odorless gas with atomic number 8, is essential for living organisms. Its melting point is $-218.4{}^{0}C$and boiling point is $-183{}^{0}C$. With electronegativity of 3.44, it is the second highly electronegative element of periodic table behind fluorine. Nitrogen is a colorless, tasteless odorless gas with atomic number 7, is the most abundant element in earth’s atmosphere. Its melting point is $-209.86{}^{0}C$and boiling point is $-195.8{}^{0}C$. With electronegativity of 3.04, it is the third highly electronegative element of the periodic table behind fluorine and oxygen.