Question

Henry's law constant for $C{{O}_{2}}$ in water is $1.67\times {{10}^{8}}Pa$ at 298 K. Calculate the quantity of $C{{O}_{2}}$ in 1L of soda water when packed under 2.5 atm $C{{O}_{2}}$ pressure at 298 K.

Answer 1

Hint: As per Henry's law, \[{{p}_{A}}=\text{ }{{K}_{H}}\chi \]
Where ${{p}_{a}}$ is partial pressure above the liquid and ${{K}_{H}}$ is Henry’s law constant.

Complete step by step answer:
> Think of a law which shows the relation between partial pressure above a liquid and amount of gas dissolved in water. The concept of mole fraction (\[\chi \]) would also come there.
> Where, \[\chi =\frac{{{n}_{A}}}{{{n}_{A}}\text{ + }{{n}_{B}}\text{ }}\text{ }\] ; A and B are two components of liquid.
> Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The proportionality factor is called Henry's law constant ${{K}_{H}}$.
Given, Henry’s law constant (${{K}_{H}}$ ) = $1.67\times {{10}^{8}}Pa$
So, for carbon dioxide gas, we can write that
${{p}_{C{{O}_{2}}}}=2.5atm=2.5\times 1.01325\times {{10}^{5}}Pa$ ($\because 1atm=1.01325\times {{10}^{5}}Pa$ )
Thus, ${{p}_{C{{O}_{2}}}}=2.5331\times {{10}^{5}}Pa$
Now, we can use the formula of Henry’s law.
\[{{p}_{A}}=\text{ }{{K}_{H}}\chi \]
So, we can write that for carbon dioxide gas as $\chi =\frac{{{p}_{C{{O}_{2}}}}}{{{K}_{H}}}$ ............(1)
Now, putting the available values into equation (1) , we get
\[\chi =\frac{2.5331\times {{10}^{5}}}{1.67\times {{10}^{8}}}\]
\[\chi =0.00152\]
Now, we can write since $C{{O}_{2}}$ is negligible as compared to water, so mole fraction is calculated by taking the value of the amount of water only.
So,
\[\text{Mole fraction of component=}\frac{\text{Moles of component}}{\text{Total moles of solution}}\]
\[\text{Mole fraction of C}{{\text{O}}_{2}}=\frac{{{n}_{C{{O}_{2}}}}}{{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}}\]
But here, ${{n}_{C{{O}_{2}}}}<<{{n}_{{{H}_{2}}O}}$
So, we can write that
 \[\text{Mole fraction of C}{{\text{O}}_{2}}=\frac{{{n}_{C{{O}_{2}}}}}{{{n}_{{{H}_{2}}O}}}\]
In 1L of soda water, volume of water = 1L
Here, we are neglecting the amount of soda present because it is very less in comparison to water.
We can write that,
1L of water = 1000 g of water
Now, $\text{Moles of water= }\frac{\text{Weight of water}}{\text{Molecular weight of water}}$
We know that Weight of water here is 1000gm and Molecular weight of water is $18gmmo{{l}^{-1}}$
$\text{Moles of water= }\frac{1000}{18}$
$\text{Moles of water= 55}\text{.55mol}$
Now, earlier we found that $\chi =0.00152$
But as $\chi $ is a mole fraction of carbon dioxide gas, we can also write that \[\chi =\frac{{{n}_{_{C{{O}_{2}}}}}}{{{n}_{_{{{H}_{2}}O}}}}\] .....(2)
Now, we will put the available values into equation (2),
\[0.00152=\frac{{{n}_{C{{O}_{2}}}}}{55.55}\]
\[{{n}_{C{{O}_{2}}}}=0.00152\times 55.55\]
\[{{n}_{C{{O}_{2}}}}=0.0844mol\]
Hence, quantity of $C{{O}_{2}}$ in 1L of soda water = Moles of $C{{O}_{2}}$ $\times $ Molar mass of $C{{O}_{2}}$
Quantity of $C{{O}_{2}}$ = 0.0844 $\times $ 44 (Molar mass of $C{{O}_{2}}$ =44$gmmo{{l}^{-1}}$ )
Quantity of $C{{O}_{2}}$ = 3.7136gm

Thus, we can conclude that the quantity of carbon dioxide gas is 3.71gm.

Note:
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.