Question

A mixture of He and Ne at a total pressure of 0.95 atm is found to contain 0.32 mol of He and 0.56 mol of Ne. The partial pressure of Ne is ________ atm.

(A) 1.7
(B) 1.5
(C) 0.60
(D) 0.35
(E) 1.0

Answer 1

The answer is C- 0.60

Step-by-step explanation:

Here, x_{Ne} = (number of moles of Ne)/(Total number of moles in mixture)

So, x_{Ne} = \frac{0.56}{0.56+0.32}=0.64

So, P_{Ne}=x_{Ne}.P_{total} = (0.64\times 0.95 atm)=0.601atm

Answer 2

Partial pressure of Ne is 0.61 atm

Step-by-step explanation:

Let's assume mixture of He and Ne behaves ideally.

Then, according to Dalton's law for partial pressure in a mixture of gas-

`P_(gas)=x_(gas).P_(total)`

Where
`P_(gas)` is partial pressure of a gas in mixture,
`x_(gas)` is mole fraction of a gas in mixture and
`P_(total)` is total pressure of mixture.

Here,
`x_(Ne)` = (number of moles of Ne)/(Total number of moles in mixture)

So,
`x_(Ne)` =
`(0.56)/(0.56+0.32)=0.64`

So,
`P_(Ne)=x_(Ne).P_(total)` =
`(0.64* 0.95 atm)=0.61atm`