Question

6. A 1.07 mg sample of a compound was dissolved in 78.1 mg of camphor. The resulting solution melted at 176.0 °C. What is the molecular mass of the compound? The melting point of pure camphor is 179.5 °C. The freezing-point-depression constant for camphor Kris 40.0 °C/m.

Answer 1

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