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

The partial pressure of ne in atm is calculated as follows

moles of Ne /total moles x total pressure

total moles 0.32 + 0.56 = 0.88 mol

partial pressure = 0.56/0.88 x0.95= 0.6045 atm

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Alleei Alleei · Answer 2 · 2020-01-02T06:04:58+01:00

Answer : The partial pressure of neon gas is, 0.61 atm

Explanation : Given,

Moles of He = 0.32 mol

Moles of Ne = 0.56 mol

Total pressure = 0.95 atm

According to the Raoult's law,

[tex]p_{Ne}=X_{Ne}\times p_T[/tex]

where,

[tex]p_{Ne}[/tex] = partial pressure of neon gas

[tex]p_T[/tex] = total pressure of gas

[tex]X_{Ne}[/tex] = mole fraction of neon gas

First we have to calculate the moles of fraction of neon gas.

[tex]\text{Mole fraction of }Ne=\frac{\text{Moles of }Ne}{\text{Moles of }Ne+\text{Moles of }He}[/tex]

[tex]\text{Mole fraction of }Ne=\frac{0.56}{0.56+0.32}=0.64[/tex]

Now we have to calculate the partial pressure of neon gas.

[tex]p_{Ne}=X_{Ne}\times p_T[/tex]

[tex]p_{Ne}=0.64\times 0.95atm[/tex]

[tex]p_{Ne}=0.61atm[/tex]

Therefore, the partial pressure of neon gas is, 0.61 atm