Question

A sample of 2.00 g of the non-volatile compound urea, CO(NH2)2 (molar mass = 60 g/mol), is dissolved in 20.0 grams of water. What is the vapor pressure of water above this solution? The vapor pressure of pure water at the temperature of the experiment is 0.150 atm. a. 0.045 atm b. 0.010 atm c. 0.038 atm d. 0.146 atm e. 0.150 atm

Answer 1

Ans: 0.146 atm

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According to raoult's law the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Since udrea is a non-volatile solid,

Psol = X(water) x Po(water)

Where,

Psol = the vapor pressure of water

Po(water) = the vapor pressure of pure water

X(water) = mole fraction of water

The mole fraction of water,

Mass of water = 20.0 g

Molar mass of water = 18.0 g/mol

No. of moles of H2O = 20.0 / 18.0 = 1.1111 mol

Mass of Urea = 2.0 g

Molar mass of urea = 60 g/mol

No. of moles of urea = 2.0 / 60 = 0.0333 mol.

The mole fraction of water is the number of moles of water divided by the total number of moles present in solution

Total number of moles of water and urea = 1.1111 mol + 0.0333 mol = 1.1444 mol

Mole fraction of water = 1.1111 mol / 1.1444 mol = 0.9709

Po(water) = 0.150 atm

Psol = X(water) x Po(water) = 0.9709 x 0.150 atm = 0.1456 atm = 0.146 atm