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

Answer:

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

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

Explanation:

$\Delta T_b=T_b-T$

$\Delta T_b=K_b\times m$

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

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

$2.63^oC=K_b\times \frac{36 g}{72 g/mol\times 0.500 kg}$

$K_b=\frac{2.63^oC\times 72 g/mol\times 0.500 kg}{36 g}=2.63 ^oC/m$

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

$0.26^oC=1\times K_b\times \frac{1.2 g}{M\times 0.050 kg}$

$M=1\times 2.63^oC/m\times \frac{1.2 g}{0.26^oC\times 0.050 kg}$

M = 242.77 g/mol

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