In a mixture of argon and hydrogen, occupying a volume of 1.18 l at 894.6 mmhg and 44.1oc, it is found that the total mass of the sample is 1.25 g. what is the partial pressure of argon?

To solve the question, there is a need to use the equation:

PV = nRT

(894.6/760) × 1.18 = n × 0.0821 × (273 + 44.1)

By solving we get:

Total moles, n = 0.053

Assume, the moles of argon as a and of hydrogen as b,

So,

40 × a + 2 × b = 1.25 (i)

a + b = 0.053 (ii)

By solving i and ii we get:

a = 0.03,

Thus, mole fraction of Ar = XAr = 0.03/0.053 = 0.57

So,

Partial pressure of Ar = 894.6 × XAr = 894.6 × 0.57 = 509.92 mm Hg

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be