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Question: In a cylinder there are \(60g\) Ne and \(64g\) Oxygen \(\left( {{O}_{2}} \right)\) . If the pressure...

In a cylinder there are 60g60g Ne and 64g64g Oxygen (O2)\left( {{O}_{2}} \right) . If the pressure of the mixture of gases in cylinder is 3030 bar then the partial pressure of O2{{O}_{2}}(in bar) in this cylinder will be
A. 30
B. 20
C. 15
D. 12

Explanation

Solution

First calculate how many moles of oxygen and Neon present in the container. The no of moles is given by the ratio of total mass of substance present and their molecular mass. After calculating the moles of Oxygen and Neon gas present in the container, calculate the total moles of gases present. And then the mole fraction of Oxygen present in the container which will be given by the ratio of moles of Oxygen present divided by the total moles of gas present. The net pressure is due to all the gases present. So the pressure exerted by the Oxygen will be the product of mole fraction of Oxygen and the total pressure.

Formulas used:
Number of moles of a substance present is given by the ratio of mass of the substance present to the molecular mass of the substance. i.e.
n=mass of the substance presentmolecular mass n=\dfrac{\text{mass of the substance present}}{\text{molecular mass }}
Mole fraction of the substance present is
f=no of moles of substancetotal moles of substancef=\dfrac{\text{no of moles of substance}}{\text{total moles of substance}}
Partial pressure is
p=mole fraction× total pressurep=\text{mole fraction}\times \text{ total pressure}

Complete step by step answer:
Given that in a cylinder there are 60g60gNe and 64g64g Oxygen(O2)\left( {{O}_{2}} \right) . If the pressure of the mixture of gases in the cylinder is 3030 bar. So
Moles of Ne present is n1=6020=3moles{{n}_{1}}=\dfrac{60}{20}=3moles
Moles of Oxygen present is n2=6432=2moles{{n}_{2}}=\dfrac{64}{32}=2moles
Mole fraction of Oxygen is
f=moles of oxygen presenttotal moles of gas=n1n1+n2=22+3=25=0.4f\text{=}\dfrac{\text{moles of oxygen present}}{\text{total moles of gas}}=\dfrac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}=\dfrac{2}{2+3}=\dfrac{2}{5}=0.4
Partial pressure exerted by Oxygen will be the product of mole fraction of oxygen and total pressure
partial pressure = mole fraction × total pressure.\text{partial pressure = mole fraction }\times \text{ total pressure}\text{.}
So the partial pressure of Oxygen is p=0.4×30=12barp=0.4\times 30=12bar
So the correct option is D.

Note:
The partial pressure is defined as the pressure exerted by the mole fraction of a gas present on a container. So the partial pressure will always be less than the total pressure. For calculation of partial pressure you have to calculate the number of moles of the substance that is present and from that calculate the mole fraction.
The sum of all the partial pressure of the gases present will be always equal to the total pressure exerted.