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Question: If \({N_2}\) gas bubbled through water at \(293K\) , how many millimoles of \({N_2}\) gas would diss...

If N2{N_2} gas bubbled through water at 293K293K , how many millimoles of N2{N_2} gas would dissolve in 1 litres1{\text{ }}litresof water? Assume that the gas exerts a partial pressure of 0.987bar0.987bar. Given that Henry’s law constant for N2{N_2} at 293K293K is 76.48Kbar76.48Kbar:
A.0.7160.716
B.0.620.62
C.0.930.93
D.0.540.54

Explanation

Solution

We have studied physical chemistry about Henry’s law. It is law that states that the amount of gas dissolved in a liquid is proportional to its partial pressure above the liquid. And the factor of proportionality is called as Henry’ law constant, which is denoted by KK.
Formula used :
P=KxP = Kx
Where, P is the pressure in atm.
K is Henry’s law constant.
x is the mole fraction

Complete step by step answer:
We should remember that, according to the henry’ law, the solubility of a gas is dependent to its mole fraction in aqueous solution.
Given data:
P=0.987barP = 0.987bar
K=76.48KbarK = 76.48Kbar (for nitrogen at 293K temperature)
We need to find the mole fraction first.
Thus,
x=PKx = \dfrac{P}{K}
Now we can substitute the known values we get,
x=0.98776480x = \dfrac{{0.987}}{{76480}}
On simplification we get,
x=1.29×10−5x = 1.29 \times {10^{ - 5}}
The term mole fraction is defined as the amount or moles of a substance divided by the total amount of the substance in a mixture.
Thus, here we already have the mole fraction and the total moles of substance that is mixture of nitrogen gas and water which is 55.5555.55 .
Hence,
n=1.29×10−5×55.55n = 1.29 \times {10^{ - 5}} \times 55.55
n=7.16moln = 7.16mol
n=0.716mmoln = 0.716mmol
Hence, the correct answer to the question is option A.

Note:
As we know that Henry's law has various applications. It is used in the production of carbonated beverages. Under high pressure, the solubility of carbon dioxide increases. We have observed that when we open the bottle of any carbonated beverage, the gas bubbles are released from the liquid. This law is also used for climbers or people living at high altitude. Because at higher altitudes, the concentration of oxygen gas is very low in the body and tissues. This leads to hypoxia. This law is also used in underwater diving.