Calculate the molar mass of a compound from the following information. you find that the freezing point of benzene is 5.5 degrees c. you then dissolve 1.60 grams of the unknown compound in 20.00 grams of benzene. the freezing point of the solution is found to be 2.8 degrees c. the freezing point depression constant for benzene is 4.3 degrees c/m.

The molar mass (Mm) of the compound is 127.39 g/mole

Explanation:

ΔT = Kf. molality

ΔT = change in temperature = Tfinal - Tinitial = 2.8 - 5.5

Kf = freezing point depression constant = - 4.3 C/m (always negative because temperature is decreasing)

molality = moles of solute/Kg of solvent = mole (n)/(20 x 10^-3 Kg of benzene)

(2.8 - 5.5) = (-4.3) x molality

molality = 0.6279 mole/kg

0.6279 = mole of compound/(20 x 10^-3)

mole of compound = 0.01256 mole

mole (n) = mass (m) divided by Molar mass (Mm)

Molar mass = mass of compound / mole of compound

m/n = 1.6/0.01256 = 127.39 g/mole

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.