Question

The melting point of biphenyl (C₁₂H₁₀) was determined to be 69.0^OC. How many grams of codeine (C₁₈H₂₁NO₃) need to be added to 16.0g of molten biphenyl to lower it's freezing point to 66.0^OC? The Kf for biphenyl is 8.00^OC*kg/mol. Report your answer without units using the correct number of significant digits.

Answer 1

Solution:

Depression on freezing point(ΔTf)=69.00 °C - 66.00 °C = 3.0 °C

Depression in freezing point is related with molar mass of the solute as,

ΔTf = 1000 x Kf x w / m x W

Where,

Kf = molal depression constant = 8.00 kg/mol

w = Mass of codeine = ?

m = molar mass of codeine = 299.364 g mol-1

W = Mass of solvent = 16 g

Therefore,

w = ΔTf x m x W / 1000 x Kf

= 3.0 °C x 299.364 g mol-1 x 16 g / 8.00 °C kg mol-1 x 1000

= 1.796 g

= 1.80 g

Therefore, mass of codeine = 1 80 g