Question

Calculate the vapor pressure of water above a solution prepared by dissolving 27.0 g of glycerin (C3H8O3) in 135 g of water at 343 K. (The vapor pressure of water at 343 K is 233.7 torr.)

Ph2o=?

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 9.00 torr at 35∘C. The vapor pressure of pure ethanol at 35∘C is 1.00×102 torr.

m=?

Answer 1

To calculate the vapor pressure of water above a solution prepared by dissolving 27.0 g of glycerin (C3H8O3) in 135 g of water at 343 K we will use the following expression:

p = p°X

here p°; The vapor pressure of water at 343 K is 233.7 torr

and X= mole fraction water

mole fraction water = moles of water / total number of moles.

moles glycerin =35.0 g/92.09 g/mol = 0.380
moles water= 130 /18.02 g/mol = 7.21
total number of moles = 7.21

mole fraction water =7.21 /7.21 + 0.380 = 0.950

p = p°X
p = 233.7 x 0.950

=222.0 torr

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 9.00 torr at 35∘C. The vapor pressure of pure ethanol at 35∘C is 1.00×102 torr.

Mass C2H5OH = 1.00 kg = 1000 g
moles C2H5OH = 1000 / 46 g/mol=21.7

p = p° X

Here p= 100-9=91 torr and p° = 100 torr and X = mole fraction of water

91 torr = 100 torr X

X = mole fraction water = 0.91

X= 0.91

0.91= 21.7 / 21.7 + moles glycol

19.75 + 0.91 moles glycol = 21.7

0.91 moles glycol = 1.953

moles glycol = 2.15 mol

than convert into mass with its molar mass:
Mass = 2.15 mol x 62 g/mol

= 133.3 g