Question

The solubility of nitrogen gas at 25°C and a nitrogen pressure of 522 mmHg is 4.7 × 10 –4 mol/L. What is the value of the Henry's Law constant in mol/L·atm?

Answer 1

Answer:

6.8 × 10⁻⁴ mol/L.atm

Explanation:

Step 1: Given data

Solubility of nitrogen gas at 25°C (S): 4.7 × 10⁻⁴ mol/L

Partial pressure of nitrogen gas (P): 522 mmHg

Step 2: Convert the partial pressure of nitrogen to atm

We will use the relationship 1 atm = 760 mmHg.

$522mmHg \times \frac{1atm}{760mmHg} = 0.687atm$

Step 3: Calculate the value of the Henry's Law constant (k)

We will use Henry's law.

$S = k \times P\\k = \frac{S}{P} = \frac{4.7 \times 10^{-4}mol/L }{0.687atm} = 6.8 \times 10^{-4} mol/L.atm$