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Question: Calculate the amount of oxygen at \[0.20{\rm{ }}atm\] dissolved in \[1{\rm{ }}kg\] of water at \[293...

Calculate the amount of oxygen at 0.20atm0.20{\rm{ }}atm dissolved in 1kg1{\rm{ }}kg of water at 293K293K. The Henry’s law constant for oxygen is 4.58×1044.58{\rm{ }} \times {\rm{ }}{10^4}atmosphere at 293K293K.
(A)7.07×10−6kg7.07{\rm{ }} \times {\rm{ }}{10^{ - 6}}kg
(B)7.27×10−6kg7.27{\rm{ }} \times {\rm{ }}{10^{ - 6}}kg
(C)7.57×10−6kg7.57{\rm{ }} \times {\rm{ }}{10^{ - 6}}kg
(D)7.77×10−6kg7.77{\rm{ }} \times {\rm{ }}{10^{ - 6}}kg

Explanation

Solution

In the given question, the solute is oxygen and the solvent is water. We have to focus on finding the Henry’s constant by using the equation as given and substituting values which are given.

Step by step answer: The real life situation in which these laws are used are to know the amount of oxygen present in the water resources to know the amount of oxygen and amount of other gases present.
Henry's law states that “At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.”
Dalton's law of partial pressures states that the partial pressure of ideal gas in a mixture of ideal gases is equal to the pressure gas would exert if it alone occupied the volume of the mixture at the temperature of the mixture.
As per Henry’s law, P=KH  xP = {K_H}\;x
where,
x is the mole fraction
KH{K_H} is the henry's law constant
P is the pressure in an atm.
Substitute values in the above expression:

x=0.20  atm4.58×104atm ⇒x=0.04×10−4x = \dfrac{{0.20\;atm}}{{4.58 \times {{10}^4}atm}}\\\ \Rightarrow x = 0.04 \times {10^{ - 4}}

We know that x=n55.55x = \dfrac{n}{{55.55}} put the value of x to find n

n=55.55  ×  0.04  × 10−4 ⇒n=2.43  ×  10−4n = 55.55\; \times \;0.04\; \times \,{10^{ - 4}}\\\ \Rightarrow n = 2.43\; \times \;{10^{ - 4}}

Thus 2.43×10−42.43 \times {10^{ - 4}} moles of oxygen is dissolved in 1L1Lof water or 1kg1kg of water.
This corresponds to 2.43×10−4  ×  32=  77.7 × 10−4gm =  7.77  ×  10−6 kg  2.43 \times {10^{ - 4}}\; \times \;32 = \;77.7\, \times \,{10^{ - 4}}gm\, = \;7.77\; \times \;{10^{ - 6}}\,kg\;of oxygen.

Hence, the correct answer is (D)    7.77×10−6kg\left( D \right)\;\;7.77{\rm{ }} \times {\rm{ }}{10^{ - 6}}kg.

Note: Here the solution is considered as an ideal solution because if it is non-ideal solution, then henry's law and Raoult’s law will be different for such non-ideal solution.