If 484 L of a gas is collected over water at 24.1°C at an atmospheric pressure of 103.2 kPa, what is the volume of the dried gas at 24.5 °C and 99.9 kPa? (The vapour pressure of water is 2 kPa.)

To solve this problem, we will use the combined gas law which states that the ratio of the product of pressure and volume to temperature of one state is equal to the ratio of the product of pressure and volume to temperature of another state. The formula is P1V1/T1 = P2V2/T2.

Step 1: Convert all temperatures to Kelvin. The Kelvin scale is used in these types of gas law problems. The conversion is K = C + 273.15. T1 = 24.1°C + 273.15 = 297.25 K T2 = 24.5°C + 273.15 = 297.65 K

Step 2: Adjust the initial pressure to account for the vapour pressure of water. The total pressure includes the pressure of the gas we are interested in and the vapour pressure of the water. We are only interested in the gas, so we subtract the water's vapour pressure from the total pressure. P1 = 103.2 kPa - 2 kPa = 101.2 kPa

Step 3: Substitute the known values into the combined gas law and solve for V2. (101.2 kPa * 484 L) / 297.25 K = (99.9 kPa * V2) / 297.65 K Solving for V2 gives V2 = 491.6 L

So, the volume of the dried gas at 24.5 °C and 99.9 kPa is approximately 491.6 L.