Question

The solubility of oxygen in water is 1.35 √ó 10‚3 mol L‚1 at 20¬∞C and 1 atm pressure. Calculate the concentration of oxygen at 20¬∞C and 0.2 atm pressure

Answer 1

According to Henry's law

Given :

P=KHxP=KHx

KH=4.34×104KH=4.34×104 atm

PO2=0.2PO2=0.2 atm

PO2=KHxO2PO2=KHxO2

∴xO2=PO2KH∴xO2=PO2KH

⇒0.24.34×104⇒0.24.34×104

⇒4.6×10−6⇒4.6×10−6

Changing mole fraction into molarity

Moles of water =100018100018

⇒55.5⇒55.5 mol

Now nO2=nO2nH2O+nO2nO2=nO2nH2O+nO2

But nO2nO2 is very small in comparision to nH2OnH2O

nO2+nH2O≈nH2OnO2+nH2O≈nH2O

∴xO2=nO2nH2O∴xO2=nO2nH2O

nO2=4.6×10−6×55.5nO2=4.6×10−6×55.5

⇒2.55×10−4⇒2.55×10−4 mol

Since 2.55×10−42.55×10−4 mol are present in 1000ml of solution,

Molarity =2.55×10−4M

Answer 2

Answer:

The concentration of oxygen at 0.2atm is 2.7*10⁻⁴ mol/L

Explanation:

Given:

Solubility(Kh)=1.35*10⁻³ mol/L

Temperature(T)=20°C=20+273=293K

Pressure(P)=1 atm

To find: Concentration of oxygen at 0.2atm

Solution:

We know that concentration is a product of solubility and pressure

It is given that solubility is equal to 1.35*10⁻³.

C=Kh*P

=0.2*1.35*10⁻³

=2.7*10⁻⁴

Therefore, the concentration of oxygen at 0.2atm is 2.7*10⁻⁴ mol/L