Question

The benzene boiling temperature (C6H6) is 80.1ºC dissolving 36 g pentane, C5H12 at 500 g benzene increases the boiling point of the solution to 82.73ºC

A. Consider the benzene boiling point constant. Show calculations.
B. In dissolving 1.2 g of unknown solute in 50 g of benzene, a solution with a boiling point of 80.36ºC is obtained, which is the molar mass of the solute (assume that i = 1) (show calculations)

Answer 1

A)Boiling point constant of benzene = 2.63°C/m

B) 242.77 g/mol is the molar mass of the solute.

Step-by-step explanation:

`\Delta T_b=T_b-T`

`\Delta T_b=K_b* m`

`Delta T_b=iK_b* \frac{\text{Mass of solute}}{\text{Molar mass of solute}* \text{Mass of solvent in Kg}}`

where,

`\Delta T_f` =Elevation in boiling point

`K_b` = boiling point constant od solvent= 3.63 °C/m

1 - van't Hoff factor

m = molality

A) Mas of solvent = 500 g = 0.500 kg

T = 80.1°C ,
`T_b` =82.73°C

`\Delta T_b=T_b-T`

`\Delta T_b`= 82.73°C - 80.1°C = 2.63°C

`2.63^oC=K_b* (36 g)/(72 g/mol* 0.500 kg)`

`K_b=(2.63^oC* 72 g/mol* 0.500 kg)/(36 g)=2.63 ^oC/m`

Boiling point constant of benzene = 2.63°C/m

B) Mass of solute = 1.2 g

Molar mas of solute = M

Mass of solvent = 50 g= 0.050 kg

i = 1

T = 80.1°C ,
`T_b` =80.36°C

`\Delta T_b=T_b-T`=80.36°C - 80.1°C = 0.26°C

`0.26^oC=1* K_b* (1.2 g)/(M* 0.050 kg)`

`M=1* 2.63^oC/m* (1.2 g)/(0.26^oC* 0.050 kg)`

M = 242.77 g/mol

242.77 g/mol is the molar mass of the solute.