Gas Volume question?

Question

Suppose a sample of refrigerant gas consisting of a simple mixture of the gases pentafluoroethane (C2HF5) and difluorormethane (CH2F2) has a density of 3.69 g/L at 20. °C and 0.873 atm.

Calculate the average molecular mass for this sample.

(This was found to be: 1.02×10^2 amu)

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The question I need some help with is:

Calculate the volume percentage of CH2F2 in the sample.

Any hints would be helpful. Thanks.

Answer 1

My answer:

Vol % = volume of solute/volume of solution or mixture * 100%. Actual density of CH2F2 is 1.1 g/mL.1 / 3.69 g/L * 1.02*10^2 amu = 27.6 L of solute. Since %vol is expresed in mL you would dilute 27.6 L to 27.6 ml and fill to 100 ml of resulting solution. Thus, .27.6 mL of solute / 100 ml of solution * 100 = 27.6 % vol wt.

Tim A.

Answer 2

It helps to be given the M.Wt. in problems like this so as to save time.

M.Wt. C2HF5 is 12.011 + 1.008 + 5*18.998 =108.009

M.Wt. CH2F2 is 12.011 + 2*1.008 + 2*18.998 = 52.023

So, say you had 22.4 liters at O centigrade and 1 atm pressure, then you would have 1 mole of particles in total, with an average mass of 102 g. Let X be the number of moles of C2HF5, then it has to be the case there are (1-x) moles of CH2F2:

102 = X * 108.009 + (1-X)*52.023
102 = (108.009 - 52.023)X + 52.023
102 - 52.023 = 55.986 * X
X = 0.8927 moles of C2HF5
this leaves 1-X = 0.1073 moles of CH2F2

Please check for errors.

Answer 3

according to the ideal gas law, one mole of gas is 22.4 liters, you know the density of gas is 3.69 g/liter. Multiply that by 22.4 liters/ mol and you get the average atomic weight of the sample. Then use the weighted average formula to calculate
(x*atomicweight1+y*atomicweight2)/(100) = average atomic weight, where x+y=100

not sure if this works but its a thought

Eseentially the second half of my answer is the same as the what is given by somone else below but I am not 100% if you calculation for average atomic mass is correct. is way is perfectly valid assuming that 102 number is correct.

Answer 4

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Answer 5

Extra data required.