(a) Give one example of a solution showing positive deviation from Raoult's law. (1 mark) (b) The vapour pressure of pure benzene at certain temperature is 0.850 bar. A non-volatile non-electrolyte solute weighing 0.5 g is added to 39 g of benzene. The vapour pressure of the solu

Question

(a) Give one example of a solution showing positive deviation from Raoult's law. (1 mark) (b) The vapour pressure of pure benzene at certain temperature is 0.850 bar. A non-volatile non-electrolyte solute weighing 0.5 g is added to 39 g of benzene. The vapour pressure of the solution is 0.845 bar. What is the molar mass of the solute added? (Given, molar mass of benzene = 78 g mol$^{-1}$) (2 marks)

**OR**

(c) What are azeotropic mixtures? (1 mark) (d) Calculate the mole fraction of ethylene glycol ($C_2H_6O_2$) in an aqueous solution containing 20% of ethylene glycol by mass. (2 marks)

Accepted Answer

>

> [!TLDR]
> (a) A classic positive-deviation pair is ethanol + acetone; (b) applying Raoult's-law relative-lowering-of-vapour-pressure to the given data gives a solute molar mass of about 169 g mol⁻¹.
>
> **(a) A solution showing positive deviation from Raoult's law**
>
> Positive deviation occurs when solute–solvent (A–B) interactions in the mixture are **weaker** than the pure-component (A–A and B–B) interactions, so molecules escape into the vapour phase more easily than Raoult's law predicts, and $\Delta_{mix}H>0$ (endothermic mixing), $\Delta_{mix}V>0$. A standard textbook example: **ethanol + acetone** (mixing partly breaks ethanol's hydrogen bonding) — other valid examples are acetone + carbon disulphide, or acetone + benzene.
>
> **(b) Molar mass of the solute**
>
> By Raoult's law, the relative lowering of vapour pressure of the solvent equals the mole fraction of solute:
> $$\dfrac{p_1^\circ - p_s}{p_1^\circ} = x_2 = \dfrac{n_2}{n_1+n_2}$$
> Given: $p_1^\circ = 0.850$ bar, $p_s = 0.845$ bar, mass of benzene $w_1=39\ g$ ($M_1=78\ g\,mol^{-1}$), mass of solute $w_2=0.5\ g$.
>
> Moles of benzene: $n_1 = \dfrac{39}{78} = 0.5\ mol$
>
> Relative lowering: $\dfrac{0.850-0.845}{0.850} = \dfrac{0.005}{0.850} = 5.88	imes10^{-3}$
>
> Since the solute is dilute ($n_2 \ll n_1$), $x_2 \approx \dfrac{n_2}{n_1}$:
> $$n_2 = x_2 	imes n_1 = 5.88	imes10^{-3} 	imes 0.5 = 2.94	imes10^{-3}\ mol$$
> $$M_2 = \dfrac{w_2}{n_2} = \dfrac{0.5}{2.94	imes10^{-3}} \approx 170\ g\,mol^{-1}$$
> (Solving the exact, non-approximated equation $x_2=n_2/(n_1+n_2)$ gives $n_2 = 2.96	imes10^{-3}\ mol$ and $M_2 \approx 169\ g\,mol^{-1}$ — the same answer to 3 significant figures.)
> …

Q.(a) Give one example of a solution showing positive deviation from Raoult's law. (1 mark)

Concept understanding — Relative Lowering of Vapour Pressure

From Intuition to a Precise Law

The Precise Statement

Why "Relative" and Why "Lowering"?

A Simple Example

Why This Works: The Concept of Relative Lowering

Step-by-Step Solution

Q.(a) Give one example of a solution showing positive deviation from Raoult's law. (1 mark)

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