Question

Calculate the solubility in water in term of mole fraction in partial pressure of $ C{{O}_{2}} $ is $ 2\times {{10}^{-3}} $ bar at $ 298K $ temperature, the $ {{K}_{a}} $ value for $ C{{O}_{2}} $ is $ 6.02\times {{10}^{-4}} $ bar.
(A) $ 3.322\times {{10}^{-3}} $
(B) $ 3.011\times {{10}^{-3}} $
(C) $ 3.322\times {{10}^{-4}} $
(D) $ 3.011\times {{10}^{-6}} $

Answer 1

Hint :We know that the answer to this question is obtained by applying Henry's law which states that the total amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas.

Complete Step By Step Answer:
In the classes of physical chemists, we have studied the concepts of various laws associated with the gases which come under the chapter called thermodynamics and some related topics. We shall now see the calculation of solubility in water in terms of mole fraction. The ratio of number of moles of the solute to that of one litre of the solution will be nothing but the molarity of the solution which is denoted as “M” and this can also be written in terms of milliliters where this doesn’t become molarity.
Henry’s law states that the total amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas. Henry’s law can be written in various forms and thus in terms of mole fraction we can write the Henry’s law in the form of equation as,
 $ {{P}_{C{{O}_{2}}}}=\left( {{K}_{a}} \right)\times \left( {{X}_{C{{O}_{2}}}} \right) $ also here we know that $ {{K}_{a}} $ is in terms of $ {{10}^{-3}} $ .
The above equation can be rewritten as; $ {{X}_{C{{O}_{2}}}}=\dfrac{{{P}_{C{{O}_{2}}}}}{{{K}_{a}}}=\dfrac{2\times {{10}^{-3}}}{6.02\times {{10}^{-4}}}\times {{10}^{-3}}=3.322\times {{10}^{-3}} $
Therefore, the correct answer is option A.

Note :
Remember that be careful with units. You have to convert them carefully so as to avoid any mistakes in the final result. Also remember the basic concepts like molarity, molality, normality, mole fraction, equivalent weight so as to solve these types of problems without any obstacle.