
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer
430.2k+ views
Hint: The partial pressure comes into significance when there is a mixture of gases. It is the pressure that is exerted by individual gas in the mixture. The total pressure is given by the sum of partial pressure of each gas in the mixture. It is called the Dalton’s Law of partial pressure.
Formula Used: ${{P}_{A}}=\text{X}{{\text{P}}_{total}}$
where ${{P}_{A}}$=Partial pressure of gas A
X=Mole fraction of gas A
${{P}_{total}}$= Total Pressure of the Mixture
Complete answer:
-First, let us consider the total mass of the gases to be 100 g. If there is presence of dihydrogen by 20% weight, then the mass of dihydrogen (H$_2$) in the mixture will be 20 g.
-So, we can calculate the mass of dioxygen (O$_2$) in the mixture i.e. 100 – 20 = 80 g.
-Now, by using the masses, let us calculate the number of moles present in the mixture;
Number of moles of dioxygen = $\dfrac{80}{32}$ = 2.5 moles; (molecular mass of O$_2$ = 32)
Number of moles of dihydrogen = $\dfrac{20}{2}$ = 10 moles; (molecular mass of H$_2$ = 2)
Total number of moles = 2.5 + 10 = 12.5 moles
-As we know,
Partial pressure of a gas = mole fraction of the gas total pressure of the mixture
Total pressure of the mixture = 1bar (given)
-Thus, partial pressure of dihydrogen = $\dfrac{no.\,of\,moles\,of\,dihydrogen}{total\,no.\,of moles}$ $\times$ total pressure
Partial pressure of dihydrogen = $\dfrac{10}{12.5}$ $\times$ 1
= 0.8 bar
So, we can conclude that partial pressure of dihydrogen is 0.8 bar.
Note: We are given with the mixture of two gases, calculate their mass, and further the number of moles can be calculated. The number of moles is attained by dividing the given mass to the molecular mass of element, or molecule. The partial pressure is the pressure exerted by individual gas, that’s why while calculating for dihydrogen there is consideration of moles of dihydrogen to total number of moles. Apply the formula of partial pressure in terms of number of moles. The partial pressure of dihydrogen could be known.
Formula Used: ${{P}_{A}}=\text{X}{{\text{P}}_{total}}$
where ${{P}_{A}}$=Partial pressure of gas A
X=Mole fraction of gas A
${{P}_{total}}$= Total Pressure of the Mixture
Complete answer:
-First, let us consider the total mass of the gases to be 100 g. If there is presence of dihydrogen by 20% weight, then the mass of dihydrogen (H$_2$) in the mixture will be 20 g.
-So, we can calculate the mass of dioxygen (O$_2$) in the mixture i.e. 100 – 20 = 80 g.
-Now, by using the masses, let us calculate the number of moles present in the mixture;
Number of moles of dioxygen = $\dfrac{80}{32}$ = 2.5 moles; (molecular mass of O$_2$ = 32)
Number of moles of dihydrogen = $\dfrac{20}{2}$ = 10 moles; (molecular mass of H$_2$ = 2)
Total number of moles = 2.5 + 10 = 12.5 moles
-As we know,
Partial pressure of a gas = mole fraction of the gas total pressure of the mixture
Total pressure of the mixture = 1bar (given)
-Thus, partial pressure of dihydrogen = $\dfrac{no.\,of\,moles\,of\,dihydrogen}{total\,no.\,of moles}$ $\times$ total pressure
Partial pressure of dihydrogen = $\dfrac{10}{12.5}$ $\times$ 1
= 0.8 bar
So, we can conclude that partial pressure of dihydrogen is 0.8 bar.
Note: We are given with the mixture of two gases, calculate their mass, and further the number of moles can be calculated. The number of moles is attained by dividing the given mass to the molecular mass of element, or molecule. The partial pressure is the pressure exerted by individual gas, that’s why while calculating for dihydrogen there is consideration of moles of dihydrogen to total number of moles. Apply the formula of partial pressure in terms of number of moles. The partial pressure of dihydrogen could be known.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How do I convert ms to kmh Give an example class 11 physics CBSE

Describe the effects of the Second World War class 11 social science CBSE

Which of the following methods is suitable for preventing class 11 chemistry CBSE
