Question

question 2 (2 points)
the solubility of o2 in 100 g of water at 10.0 °c, and a pressure of 1 atm is 0.0054 g.
the henry’s law constant for o2 at 10.0 °c and 1 atm pressure is ______ m/atm.

a) 1.7 × 10⁻⁴

b) 0.17 × 10⁻³

c) 0.05 × 10⁻⁴

d) 1.7 × 10⁻³

Explanation:

Step1: Calculate moles of O₂

Molar mass of O₂ is 32 g/mol. Moles = mass / molar mass = \( \frac{0.0054\ \text{g}}{32\ \text{g/mol}} \approx 1.6875\times10^{-4}\ \text{mol} \)

Step2: Calculate volume of water (in L)

Density of water is 1 g/mL, so 100 g water = 100 mL = 0.1 L.

Step3: Calculate molarity (M)

Molarity = moles / volume = \( \frac{1.6875\times10^{-4}\ \text{mol}}{0.1\ \text{L}} = 1.6875\times10^{-3}\ \text{M} \approx 1.7\times10^{-3}\ \text{M} \)
Henry's law: \( C = kP \), so \( k = \frac{C}{P} \). At P = 1 atm, \( k = C = 1.7\times10^{-3}\ \text{M/atm} \)

Answer:

d) \( 1.7 \times 10^{-3} \)