Question

nicotine, extracted from tobacco leaves, is a liquid completely miscible with water at temperature below 60 degrees C. if a solution made be dissolving 1.921g of nicotine in 48.92g H2O starts to freeze at -0.450 degrees C what must be the molar mass of nicotine? (kt=1.86 degrees C/m for water)

Answer 1

Given that mass of nicotine w = 1.921 g

molar mass of nicotine M = ?

mass of solvent ( water ) W = 48.92 g = 0.04892 kg

Kf = 1.86 ^oC/m

depression in freezing point dT =  Freezing point of water - Freezing point of nicotine

= 0^oC - (-0.45 ^oC)

= 0.45 ^oC

We know that

   depression in freezing point ΔT = Kf x molality

molality = (w/M) x (1/W in kg)

Then, depression in freezing point ΔT = Kf x { (w/M) x (1/W) }

M = (Kf x w ) / (ΔT x W)

=  (1.86 ^oC/m x 1.921 g) / (0.45 ^oC x  0.04892 kg)

= 162.3

M = 162.3

Therefore, molar mass of nicotine = 162.3