Q.1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol$^{-1}$. Find the molar mass of the solute.

Concept and intuition

Most liquids become denser as they cool, but water is famously anomalous: it reaches its maximum density at about 4 °C, not at its freezing point. This happens because of the balance between thermal contraction and the open, hydrogen-bonded structure that forms as water approaches freezing. For heavy water (D₂O), the stronger deuterium bonds shift this balance to a higher temperature. The difference in these maximum-density temperatures is what we need.

Step-by-step reasoning

1. Recall the known values
   • For ordinary water (H₂O), the temperature of maximum density is well known:

 In Kelvin:  

2. Find the corresponding value for heavy water (D₂O)
   • Heavy water’s maximum density occurs at a higher temperature because D₂O forms stronger hydrogen bonds (due to the greater mass of deuterium), so the open structure persists to higher temperatures.
   • The accepted experimental value is:

 In Kelvin:  
 $$
 x = 273.15 + 11.2 = 284.35\ \text{K} …

Why this approach works

The temperature of maximum density of water is a well-known physical property: for ordinary H₂O it is 3.98 °C. For heavy water (D₂O), the stronger hydrogen bonds (due to the greater mass of deuterium) shift the density maximum to a higher temperature. The difference arises from isotopic effects on molecular vibrations and hydrogen-bond strength. Instead of deriving from first principles, we recall the accepted experimental values: D₂O’s density maximum occurs near 11.2 °C. Converting to Kelvin and subtracting gives the answer.

Step-by-step reasoning

1. Recall the known value for H₂O The temperature of maximum density for ordinary water (H₂O) is $3.98\,^\circ\text{C}$. In Kelvin:

2. Recall the known value for D₂O Heavy water (D₂O) has its maximum density at about $11.2\,^\circ\text{C}$ (the exact value is often cited as 11.2 °C or 11.23 °C). In Kelvin:

3. Compute the difference $$$$

Colligative properties — like boiling point elevation, freezing point depression, and osmotic pressure — depend only on the number of solute particles in a given amount of solvent. That’s the core idea. So when you compare equimolar solutions (all at $0.1\,\text{M}$), the one that breaks into the most ions in water will have the highest effective particle concentration, and therefore the strongest colligative effect.

Let’s check each option.

1. $0.1\,\text{M}\ \mathrm{BaCl}_2$ Barium chloride dissociates fully in water:

That’s 3 ions per formula unit. So the total particle concentration is $0.1 \times 3 = 0.3\,\text{M}$.

2. $0.1\,\text{M}\ \mathrm{AgNO}_3$ Silver nitrate dissociates:

That’s 2 ions per formula unit. Particle concentration: $0.1 \times 2 = 0.2\,\text{M}$.

3. $0.1\,\text{M}\ \text{urea}$

   Urea is a covalent, non-electrolyte. It does not dissociate at all. So it remains as 1 particle per molecule. Particle concentration: $0.1\,\text{M}$.

4. $0.1\,\text{M}\ (\mathrm{NH}_4)_3\mathrm{PO}_4$

   Ammonium phosphate dissociates: …

The concept first: why does a solute lower the freezing point?

A liquid freezes at the temperature where the vapour pressure of the liquid equals the vapour pressure of the solid.

1. Dissolve a non-volatile solute in the solvent. The solute particles occupy part of the surface, so fewer solvent molecules escape — the vapour pressure of the solution falls (Raoult's law).
2. The vapour pressure of the pure solid solvent is unaffected.
3. So the two curves — solution and solid — now intersect at a lower temperature. That new intersection is the solution's freezing point.
4. Hence: the freezing point of a solution is always lower than that of the pure solvent, and the more solute you add, the lower it goes.

Because the effect depends only on the number of solute particles, not on their identity, freezing-point depression is a colligative property.

Step-by-step derivation of the graph

1. Experiment and thermodynamics both give

where $m$ is the molality and $K_f$ the cryoscopic constant (a fixed property of the solvent — for water, $K_f = 1.86\ \text{K kg mol}^{-1}$).

2. By definition,

where $T_f^{\circ}$ is the freezing point of the pure solvent and $T_f$ that of the solution.

3. Substitute:

4. Compare with the equation of a straight line, $y = c + (\text{slope})x$:
   • Variable on the $y$-axis: $T_f$
   • Variable on the $x$-axis: $m$
   • Intercept at $m = 0$: $T_f^{\circ}$ — the pure solvent's freezing point (a positive, finite value)
   • Slope: $-K_f$ — a negative constant …