Suppose a sample of refrigerant gas consisting of a simple mixture of the gases pentafluoroethane (C2HF5) and difluorormethane (CH2F2) has a density of 2.76 g/L at 37 掳C and 0.958 atm.
Calculate the volume percentage of CH2F2 in the sample
this is the full question.. the first part of the question asked to solve for the avg molecular mass for the sample (which is 73.3 amu)... i solved that part.. but im having problems solving the volume percentage of CH2F2 in the sample
if you kno the solution, details would be greatly appreciated. Thank you for your time.How do i calculate volume percentage?
Love this question.
Assume that the volume of the mixture = 1liter, x is the number of mole of C2HF5 and y is the number of mol of the second one
The mass of the mixture[ m (mixture)] = V.d = 2.76*1=2.76 (g)
The number of mole of the mixture = pV/RT = 0.0378 (mol)
m(mixture) is the sum of the mass of 2 gases :210x + 52y = 2.76
n(mixture) = x + y = 0.0378
===%26gt;x =0.005 and y=0.0328
It is due to the fact that in a mixture of gases, %V = %n
===%26gt; %V of the first one = x*100/(x+y) = 0.005*100/0.0378
= 13.2%
%V of the second one = 100% - 13,2% = 86.8%How do i calculate volume percentage?
I am assuming the given density refers to the gas mixture. Using the given density and mass (that you reported), I found the total volume of gas:
73.3g (1L gas/2.76 g) = 26.56L gas @ 37oC and 0.958 atm
Then I found what this volume would be at STP:
25.56 L gas (273K / 310 K)(0.958 atm / 1 atm) = 22.4 L gas @ STP*
* This dimensional analysis technique may be a little unusual, but it works... the volume of a gas is directly proportional to the temperature (temp goes down, volume goes down) and inversely proportional to pressure (pressure up, volume down)
Now you have the volume at STP. We know that 1 mole of any gas at STP occupies 22.4 L. Taking Y1 as the mole fraction of C2HF5 and Y2 as the mole fraction of CH2F2, Y1+Y2 = 1mole total. This should be a good start. Is there any other info given?How do i calculate volume percentage?
based on my calculations...
iluvn_q_c is correct...
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment