Question

The Henry's law constant for the solubility of helium gas in water is \(3.8 \times 10^{-4}\) M/atm at \(25^{\circ} \mathrm{C}\). (a) Express the constant for the solubility of helium gas in \(\mathrm{M} / \mathrm{mm} \mathrm{Hg}\). (b) If the partial pressure of He at \(25^{\circ} \mathrm{C}\) is \(293 \mathrm{~mm} \mathrm{Hg}\), what is the concentration of dissolved He in \(\mathrm{mol} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) ? (c) What volume of helium gas can be dissolved in \(10.00 \mathrm{~L}\) of water at \(293 \mathrm{~mm} \mathrm{Hg}\) and \(25^{\circ} \mathrm{C} ?\) (Ignore the partial pressure of water.)

Short answer

Answer: The volume of helium gas that can be dissolved in 10.00 L of water at 293 mm Hg and 25°C is approximately 6.56 x 10^{-3} L.

Step-by-step solution

Convert Henry's law constant to M/mm Hg

To convert the given Henry's law constant from M/atm to M/mm Hg, we need to use the conversion factor: 1 atm = 760 mm Hg The given constant is K = 3.8 x 10^{-4} M/atm. Now, we convert from M/atm to M/mm Hg, K (M/mm Hg) = K (M/atm) * (1 atm / 760 mm Hg) K (M/mm Hg) = (3.8 x 10^{-4}) * (1 / 760) K (M/mm Hg) = 5 x 10^{-7} M/mm Hg

Calculate the concentration of He in mol/L

To find the concentration of dissolved He given the partial pressure of He, we will apply the Henry's law: C = K * P where C is the concentration of the gas in mol/L, K is the Henry's law constant in M/mm Hg, P is the partial pressure of the gas in mm Hg. Given the partial pressure of He, P = 293 mm Hg, we can calculate the concentration of He: C = (5 x 10^{-7}) * (293) C = 1.47 x 10^{-4} mol/L

Calculate the volume of dissolved He in 10.00 L of water

To find the volume of dissolved He in the given volume of water, we will use the formula: Volume of He gas (L) = (Concentration of He (mol/L)) * (Volume of water (L)) / (Moles of He per L of gas at STP) Assuming the gas behaves ideally, the molar volume of an ideal gas at STP (Standard Temperature and Pressure, 0°C, and 1 atm) is around 22.4 L/mol. Given the concentration of He, C = 1.47 x 10^{-4} mol/L, and the volume of water, V = 10.00 L, we can find the volume of dissolved He gas: Volume of He gas = (1.47 x 10^{-4}) * (10.00) / (1/22.4) Volume of He gas = 6.56 x 10^{-3} L Therefore, the volume of helium gas that can be dissolved in 10.00 L of water at 293 mm Hg and 25°C is approximately 6.56 x 10^{-3} L.