A 9.38 gram sample of a compound is dissolved in 250.0 grams of benzene. The freezing point of this solution is 1.02 ˚C below that of pure benzene. What is the molar mass of this compound? Kf for benzene is 5.12 ˚C/m and you may assume i = 1.

THE MOLAR MASS OF THE COMPOUND IS 45.31 G/MOL.

Explanation:

Using the formula;

          Change in freezing point Δt = i Kf m

where;

i = Van Hoff constant = 1

Kf = freezing point dissociation constant = 5.12 °C /m

M = molarity = unknown

The freezing point of benzene = 5.444°C

Temperature of the final solution = 1.02°C

First is to calculate the change in the temperature

= Temperature of benzene - temperature of the solution

=  5.444 - 1.02

= 4.424 °C

Next is the use the above formula to solve for the molarity

Δt = i Kf m

4.244 = 1 * 5.12 * m / 0.250 kg of benzene

We have to divide the molarity by the mass of benzene used since it was not 1 kg if benzene that was used.

So therefore,

4.244 = 20.48 * m

m = 4.244 / 20.48

m = 0.207 mol.

Next is to calculate the molar mass of the compound:

             Molarity = mass / molar mass

Molar mass = mass/ molarity

Molar mass = 9.38 g / 0.207 mol

Molar mass = 45.31 g /mol

The molar mass of the compound is 45.31 g/mol.

the awnser is group 12