Question

1. The boiling point of ethanol, CH₃CH₂OH, is 78.500 °C at 1 atmosphere. K_b(ethanol) = 1.22 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 11.40 grams of the compound were dissolved in278.0 grams of ethanol, the solution began to boil at 78.673 °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

2. Match the following aqueous solutions with the appropriate letter from the column on the right.

	1.	0.16 m	FeI2		A.	Lowest freezing point
	2.	0.15 m	CoBr2		B.	Second lowest freezing point
	3.	0.22 m	NH4NO3		C.	Third lowest freezing point
	4.	0.40 m	Urea(nonelectrolyte)		D.	Highest freezing point

Answer 1

1)

Δ Tb = 78.673 - 78.500 = 0.17 oC

use:

Δ Tb = Kb*mb

0.173 = 1.22 *mb

mb= 0.1418 molal

m(solvent)= 278.0 g

= 0.278 kg

use:

number of mol,

n = Molality * mass of solvent in Kg

= (0.1418 mol/Kg)*(0.278 Kg)

= 3.942*10^-2 mol

mass(solute)= 11.40 g

use:

number of mol = mass / molar mass

3.942*10^-2 mol = (11.4 g)/molar mass

molar mass = 289.2 g/mol

Answer: 289.2 g/mol

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