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If solubility of vinyl chloride (g) is 0...

If solubility of vinyl chloride (g) is 0.09 M at STP then value of henry's constant will be :-

A

0.0016 bar

B

617.284 bar

C

`6.17xx10^(-2)`

D

308.642 bar

Text Solution

Verified by Experts

The correct Answer is:
B
mole fraction of vinyl chloride `=(n_(v.c))/(n_(v.c)+n_(H2O))`
`=(0.09)/(55.59)=0.00162`
`X_((v.c))=(P)/(K_(H))`
`:.K_(H)=(P)(X_(v.c))=(1)/(0.00162)=617.284` bar
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