A) 0.0263 bar
B) 69.16 bar
C) 192 bar
D) 282 bar
Correct Answer: D
Solution :
No. of moles of \[{{H}_{2}}S=0.195\] No. of moles of \[{{H}_{2}}O=\frac{1000}{18}=55.55\] Mole fraction of \[{{H}_{2}}S=~\frac{0.195}{0.195+55.55}=\text{ }0.0035\] Pressure at STP = 0.987 bar According to Henry's law, p = \[{{K}_{H}}x\] Or \[{{K}_{H}}=\frac{P{{H}_{2}}S}{X{{H}_{2}}S}=\frac{0.987}{0.0035}=282\,bar\]
You need to login to perform this action.
You will be redirected in
3 sec
