Question

71. A container of 2.461 litre at 27°C has a mixture of 0.3 mole
of N2, 0.5 mole of He and 6.2 mole of O2. What is the partial
pressure of N2?
(a) 70 atm
(b) 7 atm
(c) 23.3 atm
(d) 3 atm
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Answer 1

Answer:

The answer id D( 3 atm)

Explanation:

Let us first calculate the total pressure of a mixture.

$PV = ntotal R T P = \frac{7*0.0821*300}{2.461} Ptotal = 70.05 atm Now , Pressure exerted by N2 = \frac{number of moles of N2}{Total moles}* 70.05= \frac{0.3*70.05}{7}= 3 atm$

Answer 2

Answer:option a. 70atm

Explanation: