Question

An unopened soda can has concentration of $C{O}_2$ at ${25}^{\circ }C$ equal to 0.0506 molal. Thus, pressure of $C{O}_2$ gas in the can is:
($K_H=0.034mol/kg.bar$)

• A. 0.671 bar
• B. 1.49 bar
• C. 1.20 bar
• D. 1.71 bar

Answer 1

The correct option is B 1.49 bar

Henry Law : The solubility of a gas in a liquid at a given temperature is directly proportional to its partial pressure of the gas present above the solution.
Solubility of gas can be expressed in mole fraction based on the unit of $K_H$.
S $=K_H{P}_{C{O}_2}$
Where,
S is solubility of gas
$K_H$ is Henry's constant
$P_{C{O}_2}$ is partial pressure of the gas
$\therefore P_{C{O}_2}=\frac{\text{Solubility}}{K_H}=\frac{0.0506molk{g}^{-1}}{0.034molk{g}^{-1}ba{r}^{-1}}=1.488bar\approx 1.49bar$

Note:
If $K_H$ is in the unit of pressure, then,
$\text{Solubility}\times K_H=P$

If $K_H$ is in the unit of pressure inverse, then,
$\text{Solubility}=K_H\times P$

And solubitiy can be in terms of molarity, molality, mole fraction.