Question

If ${{N}_ {2}}$gas bubbled through water at 293 K how many millimoles of ${{N}_ {2}}$ gas would dissolve in 1 litre of water? Assume that exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for ${{N}_ {2}}$ at 293 K is 76.48 K bar:
(A) 0.716
(B) 0.62
(C) 0.93
(D) 0.54

Answer 1

Henry's law is one of the gas laws 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.

Complete step by step solution:
Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas,
We have been provided with Henry’s Law constant for $C{{O}_ {2}}$ in water is $1.67\times {{10} ^ {8}} Pa$ at 298 K,
We need to calculate the quantity of 1 L of soda water when packed under 2.5 atm pressure at 298 K,
So, as we know Henry’s Law: $P={{K}_{H}} \times x$
Where, 'P' denotes the partial pressure of the gas in the atmosphere above the liquid. 'x’ denotes the concentration of the dissolved gas.
$x=\dfrac{moles} {\dfrac {1000}{18}}$
$moles=\dfrac {0.987} {76.48\times 10} \times \dfrac {10}{18}$
So, it comes out to be: $moles=0.716\times {{10} ^ {-3}}$
So, we can say that 0.716 millimoles of ${{N}_ {2}}$ gas would dissolve in 1 litre of water,

Therefore, option (A) is correct.

Note: The main application of Henry's law in respiratory physiology is to predict how gasses will dissolve in the alveoli and bloodstream during gas exchange. The amount of oxygen that dissolves into the bloodstream is directly proportional to the partial pressure of oxygen in alveolar air.