CO2 gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO2 exerts a

Q: Why do we ignore the moles of CO2 in the denominator of the mole fraction?

The solubility of gases in liquids at normal pressures is usually very low. The moles of CO2 are negligible compared to the 50 moles of water, making the calculation much simpler without losing significant accuracy.

Q: How did we get 900 grams of water from 0.9 L?

The density of water is approximately $1\text{ g/mL}$. Since $0.9\text{ L}$ is $900\text{ mL}$, the mass is $900\text{ mL} \times 1\text{ g/mL} = 900\text{ g}$.

Q: What exactly is a millimole?

A millimole (mmol) is one-thousandth of a mole. To convert moles to millimoles, you simply multiply the number of moles by 1000.

Q: Why is Henry's constant given in 'bar'?

Henry's constant can have different units depending on how the law is expressed. When the law is $p = K_H \cdot X$, since mole fraction $X$ is unitless, $K_H$ must have the same units as pressure $p$, which is 'bar' in this case.

Q: Does temperature affect this calculation?

Yes! Henry's constant $K_H$ is highly temperature-dependent. The value $1.67 \times 10^3\text{ bar}$ is specific to $298\text{ K}$. If the temperature changes, the solubility of the gas will change.

Question

$	ext{CO}_2$ gas is bubbled through water during a soft drink manufacturing process at $298	ext{ K}$. If $	ext{CO}_2$ exerts a partial pressure of $0.835	ext{ bar}$ then $x	ext{ m mol}$ of $	ext{CO}_2$ would dissolve in $0.9	ext{ L}$ of water. The value of $x$ is ......... . (Nearest integer)
(Henry's law constant for $	ext{CO}_2$ at $298	ext{ K}$ is $1.67 	imes 10^3	ext{ bar}$)

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Fizz

Imagine you are standing in a massive soft drink manufacturing plant. The secret to that refreshing fizz in your favorite soda is carbon dioxide ($	ext{CO}_2$) gas dissolved in water. In our specific scenario, we have a container holding exactly $0.9 	ext{ L}$ of pure water at a standard room temperature of $298 	ext{ K}$.

To force the gas into the liquid, it is pressurized. The $	ext{CO}_2$ gas above the water exerts a partial pressure of $0.835 	ext{ bar}$. Our ultimate mission is to determine exactly how many millimoles of this gas successfully dissolve into the water under these conditions.

### The Master Equation: Henry's Law

How do we relate the pressure of a gas to its solubility in a liquid? This is where **Henry's Law** comes to the rescue. This elegant principle states that at a constant temperature, the partial pressure of a gas ($p$) above a liquid is directly proportional to the mole fraction ($X$) of that gas dissolved in the liquid.

Mathematically, it is expressed as:
$$p_{	ext{CO}_2} = K_H \cdot X_{	ext{CO}_2}$$

Here, $K_H$ is Henry's law constant, which is specific to the gas, the solvent, and the temperature. The problem generously provides us with $K_H = 1.67 	imes 10^3 	ext{ bar}$.

### Calculating the Mole Fraction

Let's rearrange our master equation to solve for the mole fraction of the dissolved $	ext{CO}_2$:
$$X_{	ext{CO}_2} = \frac{p_{	ext{CO}_2}}{K_H}$$

Substituting the given values:
$$X_{	ext{CO}_2} = \frac{0.835}{1.67 	imes 10^3}$$

If you look closely at the numbers, you'll notice a beautiful mathematical harmony: $1.67$ is exactly double $0.835$! This makes our calculation incredibly smooth:
$$X_{	ext{CO}_2} = \frac{1}{2 	imes 10^3} = 0.5 	imes 10^{-3}$$

This tiny number tells us a crucial physical fact: very little $	ext{CO}_2$ actually dissolves in the water under these conditions.

### The Solvent: Moles of Water

To make use of the mole fraction, we need to know the total number of moles in our solution. Let's analyze our solvent, water. We are given a volume of $0.9 	ext{ L}$, which is equivalent to $900 	ext{ mL}$.

Knowing that the density of water is approximately $1 	ext{ g/mL}$, the mass of the water is exactly $900 	ext{ g}$. To find the number of moles, we divide this mass by the molar mass of water ($18 	ext{ g/mol}$):
$$n_{	ext{H}_2	ext{O}} = \frac{900 	ext{ g}}{18 	ext{ g/mol}} = 50 	ext{ moles}$$

### The Smart Approximation

By definition, the mole fraction of $	ext{CO}_2$ is the ratio of its moles to the total moles in the solution:
$$X_{	ext{CO}_2} = \frac{n_{	ext{CO}_2}}{n_{	ext{CO}_2} + n_{	ext{H}_2	ext{O}}}$$

Here is where we apply a critical approximation. Because the solubility of $	ext{CO}_2$ is so low (as indicated by our tiny mole fraction), the number of moles of dissolved gas ($n_{	ext{CO}_2}$) is practically negligible compared to the massive $50 	ext{ moles}$ of water.

Therefore, we can safely approximate the denominator:
$$n_{	ext{CO}_2} + n_{	ext{H}_2	ext{O}} \approx n_{	ext{H}_2	ext{O}} = 50$$

This simplifies our mole fraction equation to:
$$X_{	ext{CO}_2} \approx \frac{n_{	ext{CO}_2}}{50}$$

### The Final Calculation

Now, we simply equate our two expressions for the mole fraction:
$$\frac{n_{	ext{CO}_2}}{50} = 0.5 	imes 10^{-3}$$

Multiplying both sides by $50$:
$$n_{	ext{CO}_2} = 50 	imes 0.5 	imes 10^{-3} = 25 	imes 10^{-3} 	ext{ moles}$$

The question specifically asks for the answer in **millimoles** (m mol). Since $1 	ext{ mole} = 1000 	ext{ millimoles}$, we multiply our result by $10^3$:
$$	ext{Millimoles of } 	ext{CO}_2 = 25 	imes 10^{-3} 	imes 10^3 = \mathbf{25}$$

And there we have it! Exactly $25 	ext{ millimoles}$ of carbon dioxide will dissolve to give our soft drink its perfect fizz.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Henry’s Law$

$Substituting Values$

$Calculating Mole Fraction$

$Moles of Solvent (Water)$

$Approximating Total Moles$

$Final Calculation$

The Sigma Insight: Henry's Law and Raoult's Law

The Setup

The Master Equation

Calculating the Mole Fraction

The Solvent

The Smart Approximation

The Final Calculation

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