Question

0.0840 dm³ of a gas is collected by water displacement at 22.00°C and 85.0 kPa. What volume would the dry gas occupy at standard pressure and 22.00°C? (Water vapor pressure at 22.00°C = 2.6 kPa)

A) 0.0674 dm³
B) 0.0784 dm³
C) 0.0920 dm³
D) 0.1014 dm³

Answer 1

The volume that the dry gas would occupy at standard pressure and 22.00°C is 0.0674 dm³.

Step-by-step explanation:

To find the volume that the dry gas would occupy at standard pressure and 22.00°C, we can use the ideal gas law formula: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

We need to convert the given temperature from Celsius to Kelvin: 22.00°C + 273.15 = 295.15 K.

Now we can calculate the volume of the dry gas using the formula PV = nRT. The pressure is given as 85.0 kPa, and the volume collected over water displacement is given as 0.0840 dm³. We also need to account for the water vapor pressure at 22.00°C, which is 2.6 kPa.

Substituting the values into the formula:

(85.0 kPa - 2.6 kPa) * (0.0840 dm³) / (85.0 kPa * 295.15 K) = 0.0674 dm³.

Therefore, the volume that the dry gas would occupy at standard pressure and 22.00°C is 0.0674 dm³.