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Vesna [10]
3 years ago
14

The solubility of nitrogen gas in water at 25°C and a nitrogen pressure of 522 mm Hg is 4.7 × 10⁻⁴ mol/L. What is the value of t

he Henry's law constant in mol L⁻¹• atm⁻¹?
A. 9.0 x 10⁷ mol/L atm
B. 3.2 x 10⁻⁴ mol/L atm
C. 4.7 x 10⁻⁴ mol/L atm
D. 6.8 x 10⁻⁴ mol/L atm
E. 1.5 x 10³ mol/L atm
Chemistry
1 answer:
Alona [7]3 years ago
6 0

Answer:

D. 6.8 x 10⁻⁴ mol/L atm

Explanation:

<em>Henry's Law</em> states that:

S = kH * P

Where S is the solubility of a gas (in mol/L), kH is Henry's constant (in mol• L⁻¹• atm⁻¹) and P is the partial pressure of said gas (in atm).

The problem gives us S and P, while asking us to <u>calculate kH</u>:

  • First we <u>convert 522 mmHg into atm</u>:
  • 522/760 = 0.687 atm

Now we <u>solve for kH</u>:

4.7 × 10⁻⁴ mol/L = kH * 0.687 atm

  • kH = 6.84x10⁻⁴ mol • L⁻¹• atm⁻¹
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