Question

list the following aqueous solutions in order of their expected freezing point: 0.050m CaCl2, 0.15 m HCl, 0.10 m HCl, 0.050 m CH3COOH, 0.10 m C12H22011

Answer 1

ΔTf = i*Kf*m

ΔTf is the depression in freezing point

ΔTf will be maximum when i*m is maximum

freezing temperature is maximum when ΔTb will be minimum

i is the number of ions that a compound can break into

a)

CaCl2

i = 3

CaCl2 breaks into 1 Ca2+ ion and 2 Cl- ions

so,

i*m = 3*0.050 = 0.150

b)

HCl

i=2 (1 H+ and 1 Cl-)

so,

i*m = 2*0.15 = 0.30

c)

HCl

i=2 (1 H+ and 1 Cl-)

so,

i*m = 2*0.10 = 0.20

d)

CH3COOH

i=1/2 (2 CH3COOH dimerises to form hydrogen bonding)

so,

i*m = (1/2)*0.050 = 0.0025

e)

C12H22O11

i=1

so,

i*m= 0.10

decreasing order of i*m is:

0.15 m HCl > 0.10 m HCl > CaCl2 > C12H22O11 > CH3COOH

SO, increasing order of freezing point is:

0.15 m HCl < 0.10 m HCl < CaCl2 < C12H22O11 < CH3COOH

That is freezing point is lowest for 0.15 m HCl and highest for CH3COOH