Question

1.

The boiling point of diethyl ether CH₃CH₂OCH₂CH₃ is 34.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is cholesterol .

How many grams of cholesterol, C₂₇H₄₆O (386.6 g/mol), must be dissolved in 247.0 grams of diethyl ether to raise the boiling point by 0.450 °C ? Refer to the table for the necessary boiling or freezing point constant.

Solvent	Formula	Kb (°C/m)	Kf (°C/m)
Water	H2O	0.512	1.86
Ethanol	CH3CH2OH	1.22	1.99
Chloroform	CHCl3	3.67
Benzene	C6H6	2.53	5.12
Diethyl ether	CH3CH2OCH2CH3	2.02

2.
The freezing point of benzene C₆H₆ is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is cholesterol .

How many grams of cholesterol, C₂₇H₄₆O (386.6 g/mol), must be dissolved in 258.0 grams of benzene to reduce the freezing point by 0.400°C ? Refer to the table for the necessary boiling or freezing point constant.

Solvent	Formula	Kb (°C/m)	Kf (°C/m)
Water	H2O	0.512	1.86
Ethanol	CH3CH2OH	1.22	1.99
Chloroform	CHCl3	3.67
Benzene	C6H6	2.53	5.12
Diethyl ether	CH3CH2OCH2CH3	2.02

g cholesterol.

Answer 1

Given, solvent = diethyl ether = 247.0g, boiling point Tb = 34.50°C

Solute = cholestrol , non volatile solute. Molecular weight = 386.6g/mol

Rise in boiling point of ether must be = 0.450°C

From the table, Kb of diethyl ether = 2.02 °C/m

The formula for elevation in boiling point is

ΔTb = (Kb * w*1000)/m*W

Where , w= mass of solute, m= mol wt of solute

W= weight of solvent

0.45 = (2.02*w*1000)/386.6*247.0

w= 386.6*247.0*0.45/2.02*1000

w = 21.2725 grams of cholestrol must be added.

Given, solvent= benzene= 258.0grams. Freezing point = 5.50°C

Solute= cholestrol. Non volatile solute, molecular weight= 386.6g/mol

Reduction in the freezing Point of benzene= 0.400°C

From the table, Kf of benzene = 5.12°C/m

The formula for depression in freezing point is

ΔTf = (Kf*w*1000)/m*W

0.400= 5.12*w*1000/386.6*258.0

w= 0.4*386.6*258.0/5.12*1000

w= 7.7924 grams of cholestrol must be added