Question

At what temperature in ^OC, will a 2.42msolution of ethanol in water freeze? K_fpwater=1.858^OC/m

Do NOT write units with your answer!!!

If 11.57 of an unknown covalent compound is placed in 100.0mL of water. The solution boils at 101.182^oC, what is the molar mass of the covalent compound in g/mol ?

K_bpwater=0.512^OC/m   Density of water is 1.00g/mL.

Do NOT write units with your answer!!!

If 7.29 of an unknown covalent compound is placed in 100.0mL of water. The solution boils at 100.205^oC, what is the molar mass of the covalent compound in g/mol ?

K_bpwater=0.512^OC/m   Density of water is 1.00g/mL.

Do NOT write units with your answer!!!

Choose ALL Correct:

A colligative properties depend on ...

		The concentration of solute particles.
		The identity of the solute.
		The identity of the solvent.
		Chemical reactions between the solute and solvent.

Answer 1

i)

Given data, Kf=1.858 C/mol

m=2.42

Let's Consider

$\Delta$ Tf = Kf*m

= 1.858*2.42

   $\Delta$ Tf = 4.496 degree C

ii

boiling point of water = 100 oC

boiling point of mixture = 101.182oC

dTb = 101.182 - 100

=1.18 oC

mass of unknown covalent compound = 11.57 g

density of water = 1 g/ml

mass of water = 100 ml x 1 g/ml

= 100 g

= 0.100 kg

molality of solution (m) = 11.57g/molar mass x 0.100 kg

using, dTb = iKbm

i = 1 (Van't hoff factor)

Kb for water = 0.512 oC/m

Substitute these value in above formula,

1.18=1x0.512 x 11.57/molar mass x 0.100

Molar mass of an covalent compound is 50.2 g/mol

iii)

boiling point of water = 100 oC

boiling point of mixture = 100.205oC

dTb = 100.205oC - 100

= 0.205 oC

mass of unknown covalent compound = 7.29g

density of water = 1 g/ml

mass of water = 100 ml x 1 g/ml

= 100 g

= 0.100 kg

molality of solution (m) = 7.29g/molar mass x 0.100 kg

i = 1 (Van't hoff factor)

Kb for water = 0.512 oC/m

using,

dTb = iKbm

0.205 = 1 x 0.512 x 7.29/molar mass x 0.100

Molar mass of a covalent compound is 1.82 g/mol

iv)

A colligative properties depend on The concentration of solute particles

option 1 is correct