At 25.0°C the Henry's Law constant for helium He gas in water is ×3.7/10−4Matm.Calculate the mass in grams of He gas that can be dissolved in 1475.mL of water at 25.0°C and a He partial pressure of 4.66atm.Round your answer to 2 significant digits.

To solve this problem, we need to use Henry's Law, which states that the amount of gas that dissolves in a liquid at a given temperature is proportional to the partial pressure of the gas above the liquid. The formula for Henry's Law is:

C = kP

where: C is the concentration of the gas in the liquid (in mol/L), k is the Henry's Law constant (in mol/(L*atm)), P is the partial pressure of the gas (in atm).

Given: k = 3.7 x 10^-4 mol/(L*atm), P = 4.66 atm, V = 1475 mL = 1.475 L (since 1 L = 1000 mL).

We can substitute these values into the formula to find C:

C = kP C = (3.7 x 10^-4 mol/(L*atm)) * (4.66 atm) C = 0.0017242 mol/L

This is the molar concentration of He in the water. To find the mass of He that can be dissolved, we need to convert this concentration to grams using the molar mass of He, which is approximately 4.00 g/mol:

mass = C * V * molar mass mass = (0.0017242 mol/L) * (1.475 L) * (4.00 g/mol) mass = 0.0102 g

Rounding to two significant digits, the mass of He that can be dissolved in 1475 mL of water at 25.0°C and a He partial pressure of 4.66 atm is approximately 0.010 g.