Question

The vapor pressure of ethanol (C2H5OH) at 19 ∘C is 40.0 torr. A 1.10 g sample of ethanol is placed in a 2.60 L container at 19 ∘C.

If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol remain?Express your answer to two significant figures and include the appropriate units.

Answer 1

Given,

Pressure = 40.0 Torr

Mass of sample of ethanol = 1.10 g

Volume of container = 2.60 L

Temperature = 19 °C = (19 + 273)K = 292 K

According to ideal Gas law,

pV = nRT (Here, R =  62.36 L.Torr.K-1.mol-1)

n = pV/RT = (40.00 Torr) (2.60 L)/ (62.36 L.Torr.K-1.mol-1) (292 K)

n = 0.0057mol ethanol

Molar mass of ethanol = 46.07 g/mol

Since, 1 mole of ethanol contains 46.07 g/mol

So, 0.0057 mol of ethanol will contain 0.0057 x 46.07 = 0.2626 g ≈ 0.26 g in vapor phase.

Since the total mass of ethanol taken initially is 1.10 g so, the remaining liquid ethanol = 1.10 g - 0.26 g = 0.84 g