Question

1)
What is the concentration of dissolved oxygen at 50°C under
pressure of one atmosphere if partial pressure of oxygen at
50°C is 0.14 atm.
(Henry's law constant for oxygen = 1.3x10- 'mol dm'atm-')
mula nf the alcohol that results when​

Answer 1

Answer:

9M

Explanation:

Partial pressure=Mole fraction*Total pressure

∴X$O_{2}$=0.14

X$O_{2}$=n$O_{2}$/$O_{2}$ +n$H_{2}$O

n$O_{2}$ +55.5/$nO_{2}$= 1/0.14

∴n$O_{2}$=9.0

Molarity=9M

Answer 2

The concentration of dissolved oxygen at 50 degree Celsius under pressure of 1 atmosphere if the partial pressure of oxygen at 50 degree celsius is 0.14 atm is 9 M. (Henery's law constant for oxygen = 1.3×10^-4 -mol dm atm). The step wise explanation is given below:

- It is given that the concentration of dissolved oxygen at 50 degree Celsius under pressure of 1 atmosphere if the partial pressure of oxygen at 50 degree celsius is 0.14 atm.

- It can be solved by using Henry's law.

- According to the Henry's law which is a gas law which States when the temperature is kept constant above the liquid the amount of gas that is dissolved in the liquid is directly proportional to the partial pressure of the gas.

- So, the formula of the Henry law is:

Partial pressure of substance = Henry constant (k) × concentration of substance.

- According to Henry's law

pO2 = kH × xO2 or xO2 = pO2/kH

- Given that Henry's constant for oxygen = 0.0013

xO2 = 0.14/0.0013

Mol of H2O in 1 L is 55.5

- xO2 = nO2 + 55.5/ nO2

= 1/0.14

Hence, nO2 = 9.0

- So, the molarity is 9M