Question

A mixture of CO and CO2 has vapour density 20 at STP. 100 gm of this mixture contains

Answer 1

Density = 2× 38.3 =76.6gm 100 gm of mixture = 100/76.6 mols = 1.3 mols Let the mol fraction of NO2 be X Mol wt of .

Answer 2

Hey dear,

● Answer -
0.625 moles of CO
1.875 moles of CO2

● Explanation -
# Given-
Vapor density = 20

# Solution-
Molar weight of mixture = 2 × vapor density = 2 × 20 = 40

Let
x = mole fraction of CO
1-x = mole fraction of CO2

Therefore,
28(x) + 44(1-x) = 40
28x + 44 - 44x = 40
16x = 4
x = 0.25

No of moles of CO will be -
n1 = 0.25 × 100 / 40
n1 = 0.625 moles

No of moles of CO2 will be -
n2 = (1-0.25) × 100 / 40
n2 = 1.875 moles

Therefore, 100 g of mixture contains 0.625 moles of CO and 1.875 moles of CO2.

Hope this helps...