Question

1 litre millhartre @ 100ml sooml @looome loml​

Answer 1

Answer:

Let x ml be the volume of CO  

2

​

 in the original mixture. The volume of CO will be 100−x ml.

Following reaction occurs in the presence of hot graphite.

x

CO  

2

​

(g)

​

+C(s)→  

2x

2CO(g)

​

 

The total volume of the mixture is 160 ml.

Hence, 100−x+2x=160.

Volume of CO  

2

​

 in the original mixture, x=60 ml.

Volume of CO in the original mixture =100−60=40 ml.

Volume is directly proportional to the number of moles.

Hence, mole percent of CO  

2

​

 in the mixture =  

100+40

60

​

×100=60%.

Mole fraction of CO in the mixture =  

40+60

40

​

=0.40.