Answer:
The molecular mass of the compound is 156,7 g/m
Explanation:
We have to apply the freezing-point depression formula
ΔT = kf . m
where ΔT = Melting Tº of solvent pure - Melting Tº on solution
Kf is the freezing-point-depression constant
m is molality (moles of solute in 1kg of solvent)
Let's put the values on the formula
179,5º - 176º = 40 °C/m . m
3,5ºC = 40°C/m . m
3,5ºC / 40 m/ºC = m
0,0875 m
0,0875 moles of the compound in 1kg of camphor, but you don't have 1 kg, you have just 78,1 mg, so let's convert mg into kg for the rule of three.
78,1 mg / 1x10*6 = 7,81 x10*-5 kg
So if in 1kg ___ we have ____ 0,0875 moles of the compound
in 7,81 x10*-5 kg we have .... (7,81 x10*-5 kg . 0,0875 m) / 1kg
6,83x10-6 moles of compound
The moles are the mass for 1,07 mg, but for the molar mass, we should convert the mg into g so
1,07 mg / 1000 = 1,07x10*-3 g
For the molecular mass we should divide g/m so
1,07x10*-3 g / 6,83x10-6 moles = 156,7 g/m
