Question

A student analyzes a gas mixture of sulfur hexafluoride (SF_6) and selenium hexafluoride (SeF_6) and finds that the density of the sample is 6.90 g/L at a temperature of 23 degrees C and a pressure of 0.942 atm. Calculate the partial pressure of sulfur hexafluoride.

Answer 1

D = 6.9 g/L

T = 23°C = 23+273 = 296 K

P = 0.942 atm

calcualte partial pressure of SF6

first, calculate moles of each:

PV = nRT

V/n = RT/P

V/n = (0.082)(296)/(0.942) = 25.76645 L/mol

n/V = 1/25.76645 = 0.038810 mol/L

MW of SF6 = 146.06

MW of SeF6 = 192.9534

then..

MW average = mass / mol = 6.9 g/L / 0.038810 mol / L = 6.9/0.038810= 177.78922 g /mol

MWaverage = MW-SF6*(x) + MW-SeF6(1-x)

177.78922 = 146.06 x + 192.9534 (1-x)

177.78922 = 146.06x + 192.9534 - 192.9534x

(192.95-146.06) x = 192.9534 -177.78922

x = (192.9534 -177.78922 ) / (192.95-146.06)

x = 0.323399

then...

P-SF = x-SF6 * Ptotal = 0.323399* 0.942

P-SF6 = 0.304641 atm