Question

How many grams of carbon dioxide gas is dissolved in a 1 L bottle of carbonated water if the manufacturer uses a pressure of 2.4 atm in the bottling process at $25$ $^ \circ C$?
Given- ${K_H}$ of $C{O_2}$ in water = $29.76atm{(mol/L)^{ - 1}}$ at $25$ $^ \circ C$.
(A) 3.88 g
(B) 4.90 g
(C) 5.33 g
(D) 3.52 g

Answer 1

First calculate the concentration of $C{O_2}$ in the carbonated water using Henry's Law. Then using it find the moles of $C{O_2}$ present in 1 L of carbonated water and then eventually the weight of $C{O_2}$.

Complete step by step solution:
-The dissolved carbon dioxide in carbonated drinks is an example of Henry’s Law. So, first of all we will see what Henry’s Law is.
According to Henry's Law the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid when the temperature is kept constant. Mathematically it can be written as:
$P\propto C$
$P = {K_H}C$--------------(1)
Here, ${K_H}$ is the constant of proportionality and is also known as Henry’s constant.
$P$ = Partial pressure of the gas in the atmosphere above the liquid
$C$ = Concentration of the dissolved gas.
-Coming back to the question we need to find out the weight of carbon dioxide dissolved in 1 L of carbonated water. So, first of all we will calculate the concentration of $C{O_2}$ using equation (1):
$P = {K_H}C$
The values given in the question are: $P =$ $2.4 atm$; and ${K_H}$ of $C{O_2}$ in water = $29.76atm{(mol/L)^{ - 1}}$. Putting these values in equation (1):
$\implies 2.4atm = 29.76atm{(mol/L)^{ - 1}} \times C$
$\implies C = \dfrac{{2.4atm}}{{29.76atm/(mol/L)}}$
$\implies C = 0.08mol/L$
So, we can say that $0.08$ moles of $C{O_2}$ are present in $1 L$ of the carbonated water. Also remember that the question is talking about 1 L of bottle so the number of moles of $C{O_2}$ involved is $0.08$ moles.
-Now we need to calculate the weight of $0.08$ moles of $C{O_2}$.
The molecular weight of $C{O_2}$ is $44g/mol$.
$\implies n = \dfrac{W}{M}$
$\implies 0.08 = \dfrac{W}{{44}}$
$\implies W = 3.52gm$
Finally we can say that $3.52$ grams of carbon dioxide gas is dissolved in a $1 L$ bottle of carbonated water.

Hence, the correct option will be: (D) 3.52 g

Note: Henry's Law causes the solubility of the $C{O_2}$ in the unopened drink is also high. But when we open the bottle the pressurized $C{O_2}$ gas will escape into the atmosphere because the partial pressure of $C{O_2}$ in the atmosphere above the drink will decrease and so will the solubility. As a result the dissolved $C{O_2}$ will come to the surface of the drink in the form of bubbles or effervescence to escape.