Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A gaseous mixture of He and 02 is found to have a density of 0.518 g/L at 25°C and 720 torr. What is the % by mass of helium in the mixture
First, you can use the ideal gas law to calculate the average molar mass of the gas mixture: PV = n RT Since n = mass / molar mass PV = (mass/molar mass) RT Since density = mass/V, you can rearrange the equation to give: molar mass = (density)RT / P Using the values in the problem (and converting everything to the correct units) molar mass = (0.518 g/L) (0.0821 Latm/molK) (298K) / 0.9487 atm molar mass = 13.35 g/mol Now, let x = mole fraction of He and 1-x = mole fraction of O2, Then: 13.35 g/mol = 4(x) + 32(1-x) Once you solve for x, just multiply it by 100 to get the percent of helium in the gas mixture.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -