Question

52. 600 mL of a mixture of CO and Co, weight 1 g at
NTP. What is the volume of Co, in the mixture at
NTP?
(1) 350 mL
(2) 250 mL
(3) 400 mL
(4) 200 ml​

Answer 1

Let the number of moles of carbon dioxide be x and the number of moles of carbon monoxide be y .

Now , the molecular weight of CO2 is {12 + (16*2)}

Mol Wt = {12 + 32}

Mol Wt = 44 g

So the weight of x moles of CO2 is 44*x

Weight = 44x g

The molecular weight of CO is {12 + 16}

Mol Wt = 28 g

So the weight of y moles of CO is 28*y

Weight = 28y g

The total weight of the mixture is 1 g

Therefore , 44x + 28y = 1 ….(i)

This is our first equation .

Now we know 1 mole of a gas occupies a volume of 22.4 L or 22400 mL at NTP

Therefore 1 mole of CO2 occupies 22400 mL at NTP

So x moles of CO2 occupies a volume of 22400*x at NTP

Volume = 22400x mL

Also 1 mole of CO occupies a volume of 22400 mL at NTP

So y moles of CO occupies a volume of 22400*y at NTP

Volume = 22400y mL

Now the total volume occupied by the mixture is 600 mL

Therefore , 22400x + 22400y = 600

Dividing both sides by 200 we get ,

112x + 112y = 3….(ii)

This is our second equation

If we multiply the eq(i) by 4 we get

176x + 112y = 4

112x + 112y = 3

Therefore evaluating we get , 64x = 1

x = 1/64

Therefore putting the value of x in eq(i) we get ,

44*(1/64) + 28y = 1

28y = 1 - (11/16)

28y = 5/16

y = 5/448

Therefore there are 1/64 moles of CO2 and 53/448 moles of CO

Therefore volume occupied by CO2 at STP is (1/64)*22400

Volume = 350 mL (Ans)