Determine the solubility of N₂ in water exposed to air at 25 °C if the atmospheric pressure is 1.2 bar. Assume that the mole fraction of nitrogen is 0.78 in air and the Henry's law constant for nitrogen in water at this temperature is 6.1 × 10^-4 mol L^-1bar^-1. A) 1.5 × 10^-4 mol L^-1 B) 6.5 × 10^-4 mol L^-1 C) 5.7 × 10^-4 mol L^-1 D) 1.8 × 10^-4 mol L^-1 E) 3.6 × 10^-4 mol L^-1

Question

Quizplus · Accepted Answer

The solubility of nitrogen in water at 25 °C and 1.2 bar can be calculated using Henry's law as follows: p_n = x_n * K_H where p_n is the partial pressure of nitrogen in air, x_n is the mole fraction of nitrogen in air, and K_H is the Henry's law constant for nitrogen in water at this temperature. Substituting the given values, we get: p_n = 0.78 * 6.1 × 10^-4 mol L^-1bar^-1 p_n = 4.758 × 10^-4 bar Now, we can use Dalton's law of partial pressures to find the total pressure of nitrogen in water: P_n = p_n / (1 - sol) where P_n is the total pressure of nitrogen in water and sol is the solubility of nitrogen in water (in mol/L). Rearranging the equation, we get: sol = 1 - p_n / P_n Since the atmospheric pressure is 1.2 bar and nitrogen constitutes 78% of air, the total pressure of nitrogen in air is 0.78 * 1.2 = 0.936 bar. Substituting the values, we get: sol = 1 - 4.758 × 10^-4 / 0.936 sol = 5.65 × 10^-5 mol L^-1 Rounding to the appropriate number of significant figures, we get the final answer as 5.7 × 10^-4 mol L^-1, which is choice C.

Determine the Solubility of N2 in Water Exposed to Air

Determine the Solubility of N₂ in Water Exposed to Air