Question

1. A)
The freezing point of benzene, C₆H₆, is 5.500 °C at 1 atmosphere. K_f(benzene) = 5.12 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 11.43 grams of the compound were dissolved in 267.0 grams of benzene, the solution began to freeze at 4.740 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound?

____g/mol

B) The boiling point of benzene, C₆H₆, is 80.100 °C at 1 atmosphere. K_b(benzene) = 2.53 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 10.44 grams of the compound were dissolved in294.0 grams of benzene, the solution began to boil at 80.412 °C. The compound was also found to be nonvolatile and a non-electrolyte.


What is the molecular weight they determined for this compound ?

____g/mol

Answer 1

1. A)

The freezing point of Benzene =

Freezing point of the solution =

hence, the depression in freezing point observed is

The depression in freezing point is related to the molality of the solution as follows:

Where is the proportionality constant for benzene which equals

m is the molality of the solute.

Hence, the molality m is

molality is defiene as moles of solute per kg of solvent.

Hence, 0.148 mol of solute will be in 1k g of benzene.

weight of Benzen here = 267.0 gm = 0.267 kg

Hence, number of moles of solute present in 0.267 k of Benzene is

Amount of the compound that was added = 11.43 gm

number of moles in 11.43 gm is 0.0395 mol.

hence, molar mass of the solute is

B)

The boiling point of Benzene =

Boiling point of the solution =

hence, the elevation in boiling point observed is

The elevation in boiling point is related to the molality of the solution as follows:

Where is the proportionality constant for benzene which equals

m is the molality of the solute.

Hence, the molality m is

molality is defiene as moles of solute per kg of solvent.

Hence, 0.123 mol of solute will be in 1k g of benzene.

weight of Benzen here = 294.0 gm = 0.294 kg

Hence, number of moles of solute present in 0.294 k of Benzene is

Amount of the compound that was added = 10.44 gm

number of moles in 10.44 gm is 0.0361 mol.

hence, molar mass of the solute is