Question

	_____1.	0.17 m	NiBr2		A.	Highest boiling point
	_____2.	0.24 m	NaCl		B.	Second highest boiling point
	_____3.	0.14 m	FeBr3		C.	Third highest boiling point
	_____4.	0.57 m	Sucrose(nonelectrolyte)		D.	Lowest boiling point

Answer 1

we know that

elevation in boiling point = i x Kb x m

let the normal boiling point be T

the new boling point be Tb

so

Tb-T = i x Kb x m

Tb = T + ( i x Kb x m)

now

kb is a constant

so

the boiling point depends on the product of i X Kb

higher the I x Kb value higher the boiling point

1) consider NiBr2

NiBr2 ---> Ni+2 + 2Br-

i value is the number of species in the solution after dissociation

so

in this case

i = 3

given

molality (m) = 0.17

so

i x Kb = 3 x 0.17 = 0.51

2)   now consider NaCl

NaCl ----> Na+ + Cl-

i = 2

given

m = 0.24

so

i x Kb = 2 x 0.24 = 0.48

3) consider FeBr3

FeBr3 ---> Fe+3 + 3Br-

so

i=4

given

m = 0.14

so

i x Kb = 4 x 0.14 = 0.56

4) now consider sucorse

it is a nonelectrolyte

so i value is 1

given

m = 0.57

so

i x Kb = 1 x 0.57 = 0.57

now

higher the i x Kb value higher the boiling point

so

the order is

highest sucrose > FeBr3 > NiBr2 > NaCl lowest

so

the correct matching is

1 ----> C

2----> D

3-----> B

4----> A