Question

A vessel contains equal amount of oxygen and
carbon monoxide at 10 atm. Calculate the partial
pressure of O2 under similar condition.
(1) 0.27 atm
(2) 2.7 atm
(3) 4.67 atm
(4) 5 atm​

Answer 1

Answer:

ANSWER

The mole fraction of carbon dioxide =

total number of moles of mixture

number of moles of carbon dioxide

=

0.5+0.1+0.4

0.1

=0.1

The partial pressure of carbon dioxide = mole fraction of carbon dioxide× total pressure =0.1×5=0.5 atm.

Answer 2

Final answer: (3) 4.67 atm

Given that: We are given  pressure of vessel contains equal amount of oxygen and carbon monoxide = 10 atm.

To find: We have to find partial pressure of O₂.

Explanation:

• Let "x" be the mass of each component of gas.

• Molecular mass of CO = 28g

        Molecular mass of O₂ = 32g

• Number of moles = $\frac{Given mass}{Molecular mass}$

• Total number of moles of the mixture =  $\frac{x g of CO}{28g}$  +  $\frac{x g of O2}{32g }$

            = $\frac{x}{28} + \frac{x}{32}$

            = $\frac{32x + 28x}{32*28} = \frac{60x}{896} = \frac{15x}{224}$ mole

• Partial pressure of gas = Pressure of mixture * mole fraction of gas

• Partial pressure of O₂ = $10 * [(\frac{x}{32}) / (\frac{15x}{224})]$

            $10 * \frac{224}{32 * 15} = 10 * \frac{7}{15} = 10 * 0.467$ = 4.67 atm

Option (3) is correct.

• After calculation we get partial pressure of O₂ is 4.67 atm so option (1) 0.27 atm is wrong.

• Our answer is 4.67 atm not 2.7 atm hence option (2) is wrong.

• Final answer is 4.67 atm not 5 atm hence option (4) is also wrong one.

To know more about the concept please go through the links

https://brainly.in/question/7789103

https://brainly.in/question/13523076

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