If 301 L of a gas is collected over water at 34.6°C at an atmospheric pressure of 93.7 kPa, what is the volume of the dried gas at 25.2 °C and 103.5 kPa? (The vapour pressure of water is 9.2 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 gas law problems. To convert from Celsius to Kelvin, add 273.15.

T1 = 34.6°C + 273.15 = 307.75 K T2 = 25.2°C + 273.15 = 298.35 K

Step 2: Convert all pressures to the same units. In this case, we will use kPa. The atmospheric pressure is already given in kPa, but the pressure of the gas collected over water is the atmospheric pressure minus the vapour pressure of water.

P1 = 93.7 kPa - 9.2 kPa = 84.5 kPa P2 = 103.5 kPa

Step 3: Substitute the known values into the combined gas law and solve for V2.

(V1P1)/T1 = (V2P2)/T2 (301 L * 84.5 kPa) / 307.75 K = (V2 * 103.5 kPa) / 298.35 K

Solving for V2 gives:

V2 = (301 L * 84.5 kPa * 298.35 K) / (307.75 K * 103.5 kPa) = 274.5 L

So, the volume of the dried gas at 25.2 °C and 103.5 kPa is approximately 274.5 L.