Question

a mixture of co and co2 has vapour density 20 at stp. 100 g of this mixture contains:

Answer 1

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...

Answer 2

"$From\quad the\quad given,$

$Vapour\quad density\quad of\quad { CO }_{ 2 }\quad =\quad 20$

$Molecular\quad weight\quad =\quad density\quad \times \quad 2\quad =\quad 20\quad \times \quad 2\quad =\quad 40$

$Molecular\quad weight\quad of\quad C{ O }_{ 2 }\quad =\quad 44\quad g/mol$

$Molecular\quad weight\quad of\quad CO\quad =\quad 28\quad g/mol$

$40\quad means\quad perecentage\quad of\quad C{ O }_{ 2 }\quad is\quad greater than\quad CO.$

$28(x)\quad +\quad 44(1-x)\quad =40$

$28x\quad +44-\quad 44x\quad =\quad 40$

$16x\quad =\quad 4\\ x\quad =\quad 0.25$

${ n }_{ total }\quad =\quad \frac { 100 }{ 40 } \quad =\quad 2.5$

${ n }_{ CO }\quad =\quad 25%\quad of\quad 2.5\quad =\quad \frac { 1 }{ 4 } \times \quad 2.5\quad =\quad 0.625\quad moles\quad of\quad CO.$

$100\quad g\quad ofmixture\quad contains\quad 0.625\quad moles\quad of\quad CO.$"