Question

75 g of 2-propanol (C3H8O) and 25 g of pentane are mixed in a 200 mL bottle, which is then tightly capped and stored at 20°C. Once the liquid reaches equilibrium with the gas phase, what are the partial pressures of pentane and 2-propanol in the gas? The vapor pressures at 20°C are 420 torr (pentane) and 33 torr (isopropanol).

Answer 1

Molar mass of C5H12,

MM = 5*MM(C) + 12*MM(H)

= 5*12.01 + 12*1.008

= 72.146 g/mol

Molar mass of CH8O,

MM = 1*MM(C) + 8*MM(H) + 1*MM(O)

= 1*12.01 + 8*1.008 + 1*16.0

= 36.074 g/mol

n(C5H12) = mass/molar mass

= 25.0/72.146

= 0.3465

n(CH8O) = mass/molar mass

= 75.0/36.074

= 2.079

n(C5H12),n1 = 0.3465 mol

n(CH8O),n2 = 2.079 mol

Total number of mol = n1+n2

= 0.3465 + 2.079

= 2.426 mol

use:

Mole fraction of each components are

X(C5H12) = n1/total mol

= 0.3465/2.426

= 0.1429

use:

X(CH8O) = n2/total mol

= 2.079/2.426

= 0.8571

According to Raoult’s law:

P(C5H12) = Po*X(C5H12)

P(C5H12) = 420.0 torr*0.1429

P(C5H12) = 60.0014 torr

According to Raoult’s law:

P(CH8O) = Po*X(CH8O)

P(CH8O) = 3.0 torr*0.8571

P(CH8O) = 2.5714 torr

Answer:

partial pressure of pentane = 60 torr

partial pressure of 2-propanol = 2.6 torr