Home
Class 12
CHEMISTRY
The Henry's Law constant for oxygen diss...

The Henry's Law constant for oxygen dissolved in water is `4.34xx10^(-4) Catm^(-1)` at `25^(@)C`. If partial pressure of exygen in are is 0.2 atm. Under ordinary atmospheric conditions, calculate the concentration (in moles/litre) of dissolved oxygen in water in equilibrium with air at `25^(@)C`.

Text Solution

Verified by Experts

According to Henry.s law :
`p = K_(H)x`
`K_(H) = 4.34 xx 10^(4)` atm .
`p_(O_(2)) = 0.2` atm .
`p_(O_(2)) = K_(H) x_(O_(2))`
or `x_(O_(2)) = (p_(O_(2)))/(K_(H))`
` = (0.2)/(4.34 xx 10^(4)) = 4.6 xx 10^(-6)`
Changing mole fraction into molarity .
Moles of water `= (1000)/(18) = 55.5 ` mol .
Since `n_(O_(2))` is very small in comparison to `n_(H_(2))o`
`n_(o_(2))+ n_(H_(2)o) = n_(H_(2)o)`
`x_(O_(2)) = (n_(O_(2)))/(n_(H_(2)o))`
`x_(O_(2)) xx n_(H_(2)o) = n_(O_(2))`
`n_(O_(2)) = 2.55 xx 10^(-4) mol`
since `2.55 xx 10^(-4) ` mol are present in 1000 mL of solution ,
Molarity ` = 2.55 xx 10^(-4) M`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    MODERN PUBLICATION|Exercise PRACTICE PROBLEMS|100 Videos

  • SOLUTIONS

    MODERN PUBLICATION|Exercise ADVANCED LEVEL (PROBLEMS)|21 Videos

  • SOLID STATE

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|13 Videos

  • SURFACE CHEMISTRY

    MODERN PUBLICATION|Exercise COMPETITION FILE (OBJECTIVE TYPE QUESTIONS) (MATCHING LIST TYPE QUESTIONS)|2 Videos

Similar Questions

Explore conceptually related problems

The Henry's Law constant for oxygen dissolved in water is 4.34xx10^(4) atm at 25^(@)C . If partial pressure of exygen in are is 0.2 atm. Under ordinary atmospheric conditions, calculate the concentration (in moles/litre) of dissolved oxygen in water in equilibrium with air at 25^(@)C .

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 .

The partial pressure of oxygen in air is 0.2 atm. What is the conentration of dissolved oxygen in water in equilibrium with air at 25^(@) C ? ( K_(H) for oxgyen at 25^(@)C is 4.34 cc 10^(4) atm).

The Henry's law constant for oxygen gas in water at 25^(@)C is 1.3xx10^(-3) M atm^(-1) . What is the partial pressure of O_(2) above a solution at 25^(@) C with an O_(2) concetration of 2.3xx10^(-4) M at equilibrium?

The Henry's law constant for O_(2) , dissolved in water is 4.34 xx 10^(4) atm at certain temperature. If the partial pressure of O_(2) , in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of O_(2) , in the solution?

On litre of sea water weighs 1050 grams and contains 6 xx 10^(-3)g of dissolved oxygen gas. Calculate the concentration of the dissolved oxygen in ppm.

What is the Henry's law constant of dissolved O_(2) at 10^@C at 1 atmospheric pressure , if partial pressure of oxygen is 0.24 atm ? the concentration of dissolved oxygen is 3.12xx10^(-4) mol dm ^(-3)

Applying Henry's law: The solubility of pure N_(2)(g) at 25^(@)C and 1 atm is 6.8xx10^(-4) mol L^(-1) . If the partial pressure of N_(2)(g) in the atmosphere is 0.78 atm , calculate the concentration of N_(2)(g) dissolved in water under atmospheric conditions. Strategy: According to Henry's law, the solubility of a gas in water is KP . Thus, the first step is to calculate the quantity K .