Question

The dissolution of 5.25 g of a substance in 565 g of benzene at 298 K raises the boiling point by 0.625°C. Note that Kf = 5.12 K kg/mol, Kb = 2.53 K kg/mol, and the density of benzene is 876.6 kg/m3 . Calculate the freezing point depression, the ratio of the vapor pressure above the solution to that of the pure solvent, the osmotic pressure, and the molar mass of the solute. * Pbenzene =103 Torr at 298 K.

Answer #1

Let the molar mass of the substance be 'M' g/mole

Now, moles of the substance = mass/molar mass = 5.25/M

molality of the solution = moles of the substance/mass of benzene in kg = (5.25/M)/0.565 = 9.292/M

now, elevation in B.P = K_{b}*m

or, 0.625 = 2.53*(9.292/M)

or, M = 37.614 g/mole

Now, depression in F.P = K_{f}*m = 5.12*(9.292/37.614) =
1.265 ^{0}C

density of benzene = 0.8766 g/ml

mass of solution = 570.25 g

volume of solution = mass/density = 570.25/0.8766 = 650.525 ml = 0.65053 litres

Molarity of the solution = moles of substance/volume of solution in litres = (9.292/37.612)/0.65053 = 0.38 M

Now, osmotic pressure = C*R*T

or, P = 0.38*0.0821*298 = 9.29 atm

The dissolution of 5.25 g of a substance in 565 g of benzene at
298 K raises the boiling point by 0.625 K. Note that Kf
= 5.12 K kg/mol, Kb = 2.53 K kg/mol, and the density of
benzene is 876.6 kg/m3 .
Calculate the:
a.) freezing point depression
b.) ratio of the vapor pressure above the solution to that of
the pure solvent
c.) the osmotic pressure
d.) molar mass of the solute.
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