Question

How many grams of testosterone, C19H28O2 (288.4 g/mol), must be dissolved in 299.0 grams of benzene to reduce the freezing point by 0.500°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 testosterone.

Answer 1

Answer: 8.42 grams of testosterone

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=0.500^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant  of benzene= $5.12^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 299.0 g = 0.299 kg

Molar mass of solute testosterone= 288.4 g/mol

Mass of solute testosterone added = ?

$0.500=1\times 5.12\times \frac{xg}{288.4 g/mol\times 0.299kg}$

$x=8.42g$

Thus 8.42 grams of testosterone must be dissolved in 299.0 grams of benzene to reduce the freezing point by 0.500°C.