Question

Example #2: 0.64 g of adrenaline in 36.0 g of CCl4 produces a bp elevation of 0.49 °C. What is adrenaline's molecular weight?

Solution

1) determine number of moles of adrenaline in solution:

ΔT = i Kb m
0.49 °C = (1) (4.95 °C / m) (x / 0.0360 kg)

x = 0.0035636 mol

From here it is a simple step: grams divided by moles equals the desired answer.

0.64 g / 0.0035636 mol = 180 g/mol (to two sig figs)

Answer 1

@saifoo.khan can you explain me this

Answer 2

@thomaster

Answer 3

Boiling point elevation is a colligative property. This means that it depends on how much solute there is, rather than the identity of the solute itself. The equation for boiling point elevation is:

\[\Delta T=iK_bm\]
Where:
∆T is the change in boiling point (0.49 °C here)

i is the van't Hoff factor, which is the number of dissolved particles per molecule of solute (1 in this case, since adrenaline doesn't split into more particles after dissolving)

K(b) is boiling point constant, a number that depends on the solvent being used (looks like it's 4.95 °C / m for CCl4)

m is the molality of solution, a measure of concentration that tells you the moles of solute per kilogram of solvent. The moles of solute is what we want to find (x) to eventually find molar mass, and the kilograms of solvent is 0.036 (provided in the question).

Part 1 of the solution plugs these numbers into the equation above to solve for x and gets 0.0035636 mol as the answer. That means there are this many moles of solute (adrenaline) in the solution. Part 2 of the solution uses the moles (n), along with the mass (m) of adrenaline dissolved, to find molar mass (MM). Remember, n=m/MM, so MM=m/n. Solving for MM gives 180 g/mol - the molar mass of adrenaline!

If anything is unclear please let me know!