Question

	_____1.	0.15 m	Na2S		A.	Highest boiling point
	_____2.	0.11 m	Cr2(SO4)3		B.	Second highest boiling point
	_____3.	0.17 m	MnCl2		C.	Third highest boiling point
	_____4.	0.47 m	Glucose(nonelectrolyte)		D.	Lowest boiling point

Answer 1

The boiling points of the following will be as follows,

We will consider the relation,

The equation for BP elevation is:

delta Tb = Kb x i x m

delta Tb the boiling point elevation

Kb is the ebullioscopic constant (will remain constant for all)

i = no of ions in solution

m = molality of solution

thus,

calculating for all we get,

1) 0.15 m Na2S has 3 ions

Thus,

delta Tb = 3 x 0.15 = 0.45

2) 0.11 m Cr2(SO4)3 has i = 5

Thus,

delta Tb = 5 x 0.11 = 0.55

3) 0.17 m MnCl2 has i = 3

thus,

delta Tb = 3 x 0.17 = 0.51

4) 0.47 m Glucose (non-electrolyte) has i = 1

Thus,

delta Tb = 1 x 0.47 = 0.47

From all the above calculation we can clearly say that (according to the boiling point elevation values),

1. 0.15 m Na2S will have : D lowest boiling point

2. 0.11 m Cr2(SO4)3 will have : A Highest boiling point

3. 0.17 m MnCl2 will have : B Second highest boiling point

4. 0.47 m Glucose will have : C third highest boiling point