Question

The molecular weights of sodium chloride and glucose are determined by the depression of freezing point method. Compared to their theoretical molecular weights, what will be their observed molecular weights when determined by the above method? Justify your answer.

Answer 1

The depression of freezing point method is based on colligative properties, which depend on the number of solute particles in a solution rather than the identity of the solute.

The formula used to calculate the molecular weight (M) from the depression of freezing point (ΔT_f) is

`M = (K_f * W_2 * 1000)/(Delta T_f * W_1)`

where:

K_f = freezing point depression constant

W₂ = weight of solute

ΔT_f = depression in freezing point

W₁ = weight of solvent

Sodium chloride dissociates into two ions in solution

\[\ce{NaCl -> Na+ + Cl-}\]

Therefore, for NaCl, the van’t Hoff factor (i) is 2 (since it produces two particles).

`M_"observed" = M_"theoretical"/i`

= `58.5/2`

= 29.25 g/mol

Glucose does not dissociate in solution; it remains as one molecule. Therefore, the van’t Hoff factor (i) for glucose is 1.

`M_"observed" = M_"theoretical"/i`

= `180/1`

= 180 g/mol

For sodium chloride, the observed molecular weight will be lower than the theoretical molecular weight due to dissociation into ions. 

For glucose, the observed molecular weight will be equal to the theoretical molecular weight since it does not dissociate.

Sodium Chloride (NaCl): 29.25 g/mol (observed) < 58.5 g/mol (theoretical)

Glucose (C₆H₁₂O₆): 180 g/mol (observed) = 180 g/mol (theoretical)