Home
Class 12
CHEMISTRY
Henry law constant for oxygen dissolved ...

Henry law constant for oxygen dissolved in water is` 4.34 xx 10^(4) atm` at `25^(@) C`. If the partial pressure of oxygen in air is 0.4 atm.Calculate the concentration ( in moles per litre) of the dissolved oxygen in equilbrium with air at `25^@C`.

Text Solution

Verified by Experts

Given :
Henry's law constant,`K_(H) =4.34 xx 10^(4) atm`
`P_(O_2)=0.4 atm`
According to Henry's law'
`p=K_(H)chi`
`:. pO_(2) = K_Hchio_2`
or `chio_(2)/K_(H)=0.4/4.34 xx 10^(4) =9.2 xx 10^(-6)`
Moles of water `(n_(H_(2)O)) = 1000/18 =55.5 mol`
Mole fraction of oxygen `(chiO_2)=(n_(O_2))/((n_(O_2))+n_(H_(2)O))`
Since `n_(O_2)` is very small in comparison to `n_(H_(2)O)`,
`:. chi_(O2)=n_(O_(2))/n_(H_(2)O) `
or `chi_(O_2) xx n_(H_2 O) = n_(O_2)`
`9.2 xx 10^(-6) xx 55.5 = n_(O_2)`
or `n_(O_2)=5.11 xx 10^(-4) mol`
Since `5.11 xx 10^(-4) mol` are present in `1000 mL` of solution, therefore , molarity = `5.11 xx 10^(-4)M`.
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Solved Examples|40 Videos

  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Linked Comprehension)|58 Videos

  • SOLID STATE

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 1.2 (Objective)|9 Videos

  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|2 Videos

Similar Questions

Explore conceptually related problems

The osmotic pressure of a sugar solution at 24^(@)C is 2.5 atm . The concentration of the solution in mole per litre is

Henry's law constant for the solubility of nitrogen gas in water at 298 K is 1.0 xx 10^(-5) atm . The mole fraction of nitrogen in air is 0.8 .The number of moles of nitrogen from air dissolved in 10 mol of water at 298 K and 5 atm pressure is

Henry's law constant for nitrogen at 20^@C is 76.48 k bar. Assuming the partial pressure as 0.987 bar, calculate mass of nitrogen dissovled in one litre water at 20^@C .

The Henry's law constant for the solubility of N_(2) gas in water at 298K is 1.0 xx 10^(5) atm. The mole fraction of N_(2) in air is 0.8 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298K and 5 atm pressure is

The Henry's law constant for the solubility of N_(2) gas in water at 298 K is 1.0 xx 10^(5) atm . The mole fraction of N_(2) in air is 0.8 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm . Pressure is:

Henry's law constant for oxygen and nitrogen dissolved in water at 298 K are 2.0 xx 10^(9) Pa and 5.0 xx 10^(9) Pa , respectively . A sample of water at a temperature just above 273 K was equilibrated with air (20% oxygen and 80% nitrogen ) at 1 atm. The dissolved gas was separated from a sample of this water and the dried. Determine the composition of this gas.

Henry's law constant of oxygen is 1.4 xx 10^(-3) mol "lit"^(-1) "atm"^(-1) at 298 K. How much of oxygen is dissolved in 100 mL at 298 K when the partial pressure of oxygen is 0.5 atm?

The lowest concentration of oxygen that can support aquatic life is about 1.3xx10^(-4)mol//L . The partial pressure of oxygen is 0.21 atm at sea level. What is the lowest partial pressure of oxygen that can support life ? k_(H)(O_(2))=11.3xx10^(-3)mol//L.atm

The solubility of pure oxygen in water at 20 ^(@) C and one atmosphere pressure is 1.38 xx 10 ^(-3) mol L^(-1) Calculate the concentration of oxygen at 20 ^(@) C and partial pressure of 0.21 atm.

One litre sample of sea water is found to contain 5.8 xx 10^(-3) g of dissolved oxygen. Calculate the concentration of dissolved oxygen in sea water in ppm if the density of sea water is 1.03 g cm^(-3) .