Question

Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.27 m NH4I A. Highest boiling point 2. 0.11 m FeCl3 B. Second highest boiling point 3. 0.16 m CaI2 C. Third highest boiling point 4. 0.50 m Sucrose(nonelectrolyte) D. Lowest boiling point Submit AnswerRetry Entire Group9 more group attempts remaining

Answer 1

1. 0.27 m
`NH_4I` : Highest boiling point

2. 0.11 m

: Lowest boiling point

3. 0.16 m

: Third highest boiling point

4. 0.5 m sucrose : Second highest boiling point

Step-by-step explanation:

`\Delta T_b=i* k_b* m`

`\Delta T_b` = elevation in boiling point

i = Van'T Hoff factor

`k_b` = boiling point constant

m = molality

1. For 0.27 m
`NH_4I`

`NH_4I\rightarrow NH_4^(+)+I^(-)`

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be
`2* 0.27=0.54m`

2. For 0.11 m
`FeCl_3`

`FeCl_3\rightarrow Fe^(3+)+3Cl^(-)`

i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be
`4* 0.11=0.44m`

3. For 0.16 m
`CaCl_2`

`CaCl_2\rightarrow Ca^(2+)+2Cl^(-)`

, i= 3 as it is a electrolyte and dissociate to give 3 ions, concentration of ions will be
`3* 0.16=0.48m`

4. For 0.5 m sucrose

, i= 1 as it is a non electrolyte and does not dissociate to give ions, concentration will be 0.5 m

Thus as concentration of solute follows the order :
`NH_4I` > sucrose >
`CaCl_2` >
`FeCl_3`, the boiling point will also follow the same order.