Answer:
The right solution is "[tex]6.55\times 10^{-3} \ g/100 \ ml[/tex]"
Explanation:
The given values are:
Total pressure,
= 4.79 atm
Mole fraction,
= 78.1
Solubility,
S = [tex]1.75\times 10^{-3}/100[/tex]
Partial pressure,
P = 1
By using the Henry's law,
⇒ [tex]S=K\times P[/tex]
On putting the given values, we get
⇒ [tex]\frac{1.75\times 10^{-3}}{100} = K\times 1[/tex]
⇒ [tex]K=1.75\times 10^{-5} g/ml \ atm[/tex] (Henry's constant)
The pressure of nitrogen ([tex]P'[/tex]) will be:
= [tex]Total \ pressure\times Mole \ fraction[/tex]
On substituting the above given values, we get
= [tex]\frac{4.79\times 78.1 }{100}[/tex]
= [tex]3.741 \ atm[/tex]
New solubility of nitrogen will be:
⇒ [tex]S' = K\times P'[/tex]
⇒ [tex]=1.75\times 10^{-5}\times 3.741[/tex]
⇒ [tex]=6.55\times 10^{-5} \ g/ml[/tex]
So,
`The solubility of water will be:
= [tex]6.55\times 10^{-5}\times 100[/tex]
= [tex]6.55\times 10^{-3} \ g/100 \ ml[/tex]
