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.

8.42 grams of testosterone

Explanation:

Depression in freezing point is given by:

= Depression in freezing point

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

= freezing point constant  of benzene=

m= molality

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

Molar mass of solute testosterone= 288.4 g/mol

Mass of solute testosterone added = ?

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

a is the answer

explanation:

which statement about trans-fatty acids is false?

a. the conversion of cis-fatty acids to trans-fatty acids is inhibited by the presence of antioxidants.

hope this

the correct answer to this question is 2