Question

1. Put in order from lowest to highest freezing point

0.21 m AgNO3, 0.13 M Zn(CH3COO)2, 0.15 M NiI2, 0.48 M urea(non electrolyte)

2.Put in order from lowest to highest freezing point

0.19 m CoBr2, 0.13 m AlCl3, 0.28 m ZnSO4, 0.49 m Glucose( non electrolyte)

3. Put in order from lowest to highest boiling point

0.26 m KOH, 0.16 m BaI2, 0.23 m CuSO4, 0.43 m Sucrose(nonelectrolyte)

4.Put in order from lowest to highest boiling point

0.12 m FeBr2, 0.16 m CrBr2, 8.6 x 10^-2 m Cr2(SO4)3, 0.52 m Urea( nonelectrolyte)

Answer 1

Colligative property is dependent on no of solute particles i.e. no of ions/species.

Species with highest value of (molality or molarity x no of ions/species) has highest depression in freezing pont which in turn has lowest freezing point. ( Depression in freezing pont is a colligative property )

Species with highest value of (molality or molarity x no of ions/species) has highest elevation in boiling pont which in turn has highest boiling point. ( Elevation in boiling point is a colligative property )

Non-electrolytes won't produce any ions.

1) 0.21 m AgNO3, 0.13 M Zn(CH3COO)2, 0.15 M NiI2, 0.48 M urea(non electrolyte)

a) 0.21 m AgNO3

= 0.21 x 2 ( AgNO3 = Ag+ + NO3- = 2 ions)

= 0.42

  b) 0.13 M Zn(CH3COO)2

= 0.13 x3 [ Zn(CH3COO)2 = Zn2+ + 2CH3COO- = 3 ions]

= 0.39

c) 0.15 M NiI2

= 0.15 x3 [ NiI2 = Ni2+ + 2I- = 3 ions ]

= 0.45

d) 0.48 M urea (non electrolyte)

= 0.48 x1

= 0.48

Hence , order of freezing points is

0.48 M urea < 0.15 M NiI2 < 0.21 m AgNO3 <0.13 M Zn(CH3COO)2

lowest highest

2) 0.19 m CoBr2, 0.13 m AlCl3, 0.28 m ZnSO4, 0.49 m Glucose( non electrolyte)

a) 0.19 m CoBr2

= 0.19 x 3 ( CoBr2 = Co2+ + 2Br- = 3 ions)

= 0.57

  b) 0.13 m AlCl3

= 0.13 x 4 [  AlCl3 = Al3+ + 3Cl- = 4 ions]

= 0.52

c) 0.28 m ZnSO4

= 0.28 x2 [ ZnSO4 = Zn2+ + SO42- = 2 ions ]

= 0.56

d) 0.49 M Glucose (non electrolyte)

= 0.49 x1

= 0.49

Hence , order of freezing points is

0.19 m CoBr2 < 0.28 m ZnSO4< 0.13 m AlCl3 < 0.49 m Glucose

lowest highest

c) 0.26 m KOH, 0.16 m BaI2, 0.23 m CuSO4, 0.43 m Sucrose (nonelectrolyte)

a) 0.26 m KOH

= 0.26 x 2 ( KOH = K+ + OH- = 2 ions)

= 0.52

  b) 0.16 m BaI2

= 0.16 x 3 [BaI2 = Ba2+ + 2I- ]

= 0.48

c) 0.23 m CuSO4

= 0.23 x2 [ CuSO4 = Cu2+ + SO42- = 2 ions ]

= 0.46

d) 0.43 m Sucrose (nonelectrolyte)

= 0.43 x1

= 0.43

Hence , order of boiling points is

0.43 m Sucrose < 0.23 m CuSO4 < 0.16 m BaI2 < 0.26 m KOH

lowest highest

4) 0.12 m FeBr2, 0.16 m CrBr2, 8.6 x 10^-2 m Cr2(SO4)3, 0.52 m Urea( nonelectrolyte)

a) 0.12 m FeBr2

= 0.12 x 3 ( FeBr2 = Fe2+ + 2Br- = 3 ions)

= 0.36

  b) 0.16 m CrBr2

= 0.16 x 3 [CrBr2 = Cr2+ + 2Br- ]

= 0.48

c) 8.6 x 10^-2 m Cr2(SO4)3

= 0.086 x 5 [ Cr2(SO4)3 = 2Cr3+ + 3SO42- = 5 ions ]

= 0.43

d) 0.52 m Urea( nonelectrolyte)

= 0.52 x1

= 0.52

Hence , order of boiling points is

0.12 m FeBr2 <8.6 x 10^-2 m Cr2(SO4)3 <  0.16 m CrBr2 < 0.52 m Urea( nonelectrolyte)

lowest highest