Question

A) The normal boiling point of ethanol, CH3CH2OH, is 78.4 ∘C. When 8.45 g of a soluble nonelectrolyte is dissolved in 100.0 g of ethanol at that temperature, the vapor pressure of the solution is 7.40×102torr. What is the molar mass of the solute?

Answer 1

P_solution = X_ethanol .P°_ethanol (1)

According to the definition of boiling point, at boiling point the vapour pressure becomes equal to the atmospheric pressure.

Atmospheric pressure = 1 atm = 760 torr = vapour pressure of pure ethanol (solvent) = P^°_ethanol

P_solution = 7.40 × 10² torr = 740 torr

Putting the value in equation (1), we get

740 torr = X_ethanol . 760 torr

X_ethanol = 0.9737 = n_ethanol/n_ethanol+ n_{solute (nonelectrolytes)} (2)

n_ethanol = mass/ molar mass

= 100.0 g /46 g mol^-1

= 2.17 mol

Putting the in equation (2), we get

0.9737 = 2.17 mol/2.17 mol + n_solute

or n_solute = 0.0586 mol = mass/ molar mass

= 8.45 g/molar mass

or Molar mass = 8.45g/0.0586 mol

= 144.2 g/mol ( answer)