Q.If $N_2$ gas is bubbled through water at 293 K, how many millimoles of $N_2$ gas would dissolve in 1 litre of water? Assume that $N_2$ exerts a partial pressure of 0.987 bar. Given that Henry's law constant for $N_2$ at 293 K is 76.48 kbar.

Concept: Henry's Law — the mole fraction of a gas dissolved in a liquid is directly proportional to its partial pressure: $p = K_H \cdot x$.

Step 1 — Write Henry's law for mole fraction:

Given $p_{N_2} = 0.987\ \text{bar}$ and $K_H = 76.48\ \text{kbar} = 76480\ \text{bar}$.

Step 2 — Calculate $x_{N_2}$:

Step 3 — For 1 L of water, moles of water $= \frac{1000\ \text{g}}{18\ \text{g/mol}} = 55.5\ \text{mol}$ (the value NCERT uses). Since $x_{N_2}$ is very small: …

Why Henry's law works here

When a gas is bubbled through water, it dissolves until the pressure of the gas above the liquid equals the partial pressure in the bubble. Henry's law tells us that the concentration of a dissolved gas is directly proportional to its partial pressure above the solution:

where $p$ is the partial pressure of the gas, $k_H$ is Henry's law constant, and $x$ is the mole fraction of the gas in the solution. The constant $k_H$ is given in kbar, so we must be careful with units.

The key insight: we are asked for millimoles in 1 litre of water. That means we need the mole fraction first, then convert it to moles using the amount of water.

Step-by-step solution

1. Write Henry's law and identify the knowns

Given:

• $p = 0.987$ bar
• $k_H = 76.48$ kbar = $76.48 \times 10^3$ bar (since 1 kbar = 1000 bar)

So:

2. Calculate the mole fraction

This tiny number makes sense — gases are sparingly soluble in water.

3. Relate mole fraction to moles of N₂

   For a dilute solution, the mole fraction of N₂ is:

Since $n_{N_2}$ is very small compared to $n_{H_2O}$, we can approximate:

This approximation is excellent here because $n_{N_2} \ll n_{H_2O}$.

4. Find moles of water in 1 litre

   Density of water at 293 K ≈ 1 g/mL, so 1 L = 1000 g.

   Taking the molar mass of water as 18 g/mol (as NCERT does):

5. Solve for moles of N₂ …

Method: Henry's Law (Mole Fraction → Moles in Solution)

Concept: Henry's Law states that at constant temperature, the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid.

Steps

Step 1 — Write Henry's Law

For a gas dissolved in a liquid:

where:

• $p$ = partial pressure of the gas above the liquid (in bar)
• $K_H$ = Henry's law constant (in bar)
• $x$ = mole fraction of the gas in the solution

Step 2 — Convert given data to consistent units

Given:

• $p = 0.987 \text{ bar}$
• $K_H = 76.48 \text{ kbar} = 76.48 \times 10^3 \text{ bar}$

Step 3 — Find mole fraction of $N_2$

Step 4 — Relate mole fraction to moles

For 1 litre of water:

• Mass of water = $1000 \text{ g}$ (since density ≈ 1 g/mL)
• Moles of water = $\frac{1000}{18} = 55.5 \text{ mol}$ (the value NCERT uses)

Mole fraction of $N_2$: …

Common Mistakes in Henry's Law Problems (N₂ in Water)

Students often lose marks on this exact type of problem. Here are the most frequent errors and how to avoid each:

✗ Mistake 1: Forgetting to convert units (kbar → bar)

The error:

Henry's constant is given as 76.48 kbar, but partial pressure is in bar. Students plug in $K_H = 76.48$ directly without converting.

Why it's wrong:

Henry's law:

If $K_H$ is in kbar and $P$ is in bar, the units don't match — the answer will be off by a factor of 1000.

✓ How to avoid:

Always check units before substituting. Convert:

✗ Mistake 2: Confusing mole fraction with moles dissolved

The error:

Students take $x$ (mole fraction) as the number of moles of N₂ dissolved.

Why it's wrong:

$x$ is a ratio — it tells you the fraction of total moles that are N₂. To get actual moles of N₂, you need:

✓ How to avoid:

Remember:

For dilute solutions, $n_{\text{water}} \gg n_{N_2}$, so:

✗ Mistake 3: Using wrong molar mass or volume of water

The error:

Using 1 L water = 1000 g, but then fumbling the conversion of grams to moles, or using an inconsistent molar mass.

Why it's wrong:

The standard value here is 18 g/mol. The bigger error is forgetting to convert grams to moles at all.

✓ How to avoid:

For 1 L water (≈ 1000 g):

(This is the value NCERT's Solution uses.)

✗ Mistake 4: Stopping at mole fraction — not converting to millimoles

The error:

After finding $n_{N_2}$ in moles, students forget to convert to millimoles.

Why it's wrong: …