Answer to Question #86553 in General Chemistry for mike

Question

The partial pressure of water vapor in saturated air at 31 °C is 4.43×10-2 atm.

1. How many molecules of water are in 5.83 cm3 of saturated air at 31 °C?

_________ Molecules.
2. What volume of saturated air at 31 °C contains 0.559 mol of water?
_________ L.

Answer 1

Answer on Question #86553 – Chemistry – General Chemistry

Task:

The partial pressure of water vapor in saturated air at 31 °C is 4.43×10^-2 atm.

1. How many molecules of water are in 5.83 cm3 of saturated air at 31°C?

_________ Molecules.

2. What volume of saturated air at 31°C contains 0.559 mol of water

_________ L.

Solution (1):

According to ideal gas equation: PV = nRT

Pressure = P = 4.43*10^-2 atm = 0.0443 atm.

Volume = V = 5.83 cm³ = 5.83 mL = 0.00583 L.

Temperature = T = 31°C = (31 + 273.15) = 304.15 K.

Gas constant = R = 0.08206 L*atm*mol^-1*K^-1.

Then,

n(H₂O, g) = PV / RT = (0.0443 atm * 0.00583 L) / (0.08206 L*atm*mol^-1*K^-1 * 304.15 K) = 1.035*10^-5 mol.

N = n*Na, where N = number of molecules, n = number of moles in substance, and N_a is Alvogadro’s constant.

Then,

N(H₂O, g) = n(H₂O, g) * N_a = 1.035*10^-5 mol * 6.022*10²³ mol^-1 = 6.2328*10¹⁸ molecules.

N(H₂O, g)  = 6.2328*10¹⁸ molecules.

Answer (1): 6.2328*10¹⁸ molecules of water. 

Solution (2):

According to ideal gas equation: PV = nRT

V(H₂O, g)  = nRT / P = (0.559 mol * 0.08206 L*atm*mol^-1*K^-1 * 304.15 K) / (0.0443 atm) = 314.9397 L = 314.94 L

V(H₂O, g) = 314.94 L.

Answer(2): 314.94 L of saturated air.