Question

Supposed that 6.0 g of a mixture of naphthalene and anthracene is dissolved in 300 g of benzene. When the solution is cooled it begins to freeze at a temperature of 0.70 C below the freezing point of pure benzene (5.5 C). Find the composition of themixture given that Kf for benzene is 5.1 C kg/mol.

Answer 1

wwe have formula

dT = i x Kf x m           

where dT = freezing point of solvent - freezing point of solution = 0.7 C

i = vantoff factor = 1 for non dissociate solute

Kf = 5.1 C/kgmol

hence 0.7 C = 1 x 5.1 C/Kgmol x m

m = 0.137255 = moles of solute / solvent mass in kg        where solvent mass = 300g = 0.3 kg

0.137255 = ( moles of solute ) / ( 0.3 kg)

moles of solute = 0.04118            

Let napthalene mass be   X , anthracene mass = 6-X

napthalene moles = mass / Molar mass of napthalene = X g / (128.17 g/mol) = 0.0078 X

anthracene moles = mass / molarmass of anthracene = (6-X) / ( 178.23 g/mol) = 0.0056 ( 6-X)

now 0.0078X + ( 0.0056) ( 6-X) = 0.04118

0.0022 X   = 0.00758

X = 3.45 g

thus napthalene mass = 3.45 g, anthracene mass = 2.55 g