Question

Which of the following solutions has the lowest boiling point? EXPLAIN.

A)	0.15 M NaCl
B)	0.10 M MgBr2
C)	0.15 M Ba(NO3)2
D)	0.20 M C2H6O2
E)	0.10 M Fe(NO3)3

Answer 1

Tb   = i*m*Kb

   NaCl --------------> Na^+ (aq) + Cl^- (aq)

           i = 2

m   = 0.15M

Kb   = 0.512⁰C/m

Tb   = 2*0.15*0.512    = 0.1536⁰C

The boiling point   = 100+0.1536    = 100.1536⁰C

b.

Tb   = i*m*Kb

MgBr2 --------------------> Mg^2+ (aq) + 2Br^-

     i = 3

m   = 0.10M

Kb   = 0.512⁰C/m

Tb   = 3*0.10*0.512    = 0.1536⁰C

The boiling point   = 100+0.1536    = 100.1536⁰C

c.

Tb   = i*m*Kb

Ba(NO3)2 -------------------> Ba^2+ (aq) + 2NO3^- (aq)

i = 3

m   = 0.15M

Kb   = 0.512⁰C/m

Tb   = 3*0.15*0.512    = 0.2304⁰C

The boiling point   = 100+0.2304 = 100.2304⁰C

d.

Tb   = i*m*Kb

non electrolyte i= 1

m   = 0.20M

Kb   = 0.512⁰C/m

Tb   = 1*0.15*0.512    = 0.0768⁰C

The boiling point   = 100+0.0768 = 100.0768⁰C >>>>lowest boiling point

D. 0.20 M C₂H₆O₂      lowest boiling point >>>>answer

e.

Tb   = i*m*Kb

Fe(NO3)3 -------------------> Fe^3+ (aq) + 3NO3^- (aq)

                        i = 4

m   = 0.10M

Kb   = 0.512⁰C/m

Tb   = 4*0.1*0.512    = 0.2048⁰C

The boiling point   = 100+0.2048 = 100.2048⁰C