Question

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3. 0.13 m Mn(NO3)2 4. 0.31 m Sucrose(nonelectrolyte) A. Highest boiling pointB. Second highest boiling pointC. Third highest boiling point D. Lowest boiling point

Answer 1

0.13 m of
`Mn(NO_3)_2` → Highest boiling point

0.19 m of
`AgNO_3` → Second Highest boiling point

0.17 m of
`CrSO_4` → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Step-by-step explanation:

Elevation in boiling is given by :

`\Delta T_b=i* k_b* m`

Where :

i = van't Hoff factor

`k_b`= Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of
`AgNO_3`

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

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

`\Delta T_b=2* k_b* 0.19 m`

`\Delta T_b=0.38 m* k_b`

2) 0.17 m of
`CrSO_4`

`CrSO_4\rightarrow Cr^(2+)+SO_4^(2-)`

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

`\Delta T_b=2* k_b* 0.17 m`

`\Delta T_b=0.34 m* k_b`

3) 0.13 m of
`Mn(NO_3)_2`

`Mn(NO_3)_2\rightarrow Mn^(2+)+2NO_3^(-)`

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

`\Delta T_b=3* k_b* 0.13 m`

`\Delta T_b=0.39 m* k_b`

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

`\Delta T_b=1* k_b* 0.31 m`

`\Delta T_b=0.31 m* k_b`

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

`0.39 m* k_b>0.38 m* k_b>0.34 m* k_b>0.31 m* k_b`

0.13 m of
`Mn(NO_3)_2` → Highest boiling point

0.19 m of
`AgNO_3` → Second Highest boiling point

0.17 m of
`CrSO_4` → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point