Question

Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.11 m FeBr3 A. Highest boiling point 2. 0.15 m CuBr2 B. Second highest boiling point 3. 0.24 m AgNO3 C. Third highest boiling point d 4. 0.51 m Glucose(nonelectrolyte) D. Lowest boiling point

Answer 1

1. For 0.11 m
`FeBr_3` : Lowest boiling point

2. For 0.15
`CuBr_2`

: Second highest boiling point

3. For 0.24
`AgNO_3`

: Third highest boiling point

4. 0.51 m glucose : Highest boiling point

Step-by-step explanation:

Elevation in boiling point:

`\Delta T_b=ik_b* m`

where,

`T_b` = change in boiling point

i= vant hoff factor

`k_b` = boiling point constant

m = molality

1. For 0.11 m
`FeBr_3`

`FeBr_3\rightarrow Fe^(3+)+3Br^(-)`

, i= 4 as it is a electrolyte and dissociate to give 4 ions and concentration of ions will be
`1* 0.11+3* 0.11=0.44`

2. For 0.15
`CuBr_2`

`CuBr_2\rightarrow Cu^(2+)+2Br^(-)`

, i= 3 as it is a electrolyte and dissociate to give 3 ions, concentration of ions will be
`1* 0.15+2* 0.15=0.45`

3. For 0.24
`AgNO_3`

`AgNO_3\rightarrow Ag^(+)+NO_3^(-)`

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be
`1* 0.24+1* 0.24=0.48`

4. 0.51 m glucose

i= 1 as it is a non electrolyte and does not dissociate to give ions, concentration will be
`1* 0.51=0.51`

Thus as boiling point depends on the concentration of solutes, the solution having highest concentration will have highest boiling point.