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

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* 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* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \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* 5.12* (xg)/(288.4 g/mol* 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.