Gujarat BoardEnglish MediumSTD 12 ScienceChemistrySolutions4 Marks
Question
Write Raoult's Law for Non-Volatile solute and volatile solvent and derive it's formula.
✓
Answer
→ The vapour pressure of a solvent in solution is less than that of the pure solvent. → Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity. → A relation between vapour pressure of the solution, mole fraction and vapour pressure of the solvent, $p_1=p_1^0 \cdot x_1$ → The reduction in the vapour pressure of solvent $\left(\Delta p_1\right)$ is given as : $ \begin{aligned} \Delta p_1 & =p_1^0-p_1 \\ \therefore \Delta p_1 & =p_1^0-p_1^0 \cdot x_1 \\ \therefore \Delta p_1 & =p_1^0\left(1-x_1\right) \\ \therefore \Delta p_1 & =p_1^0 x_2 \end{aligned} $ → The lowering of the vapour pressure is directly proportional to mole-Fraction of solute. $\begin{aligned} \therefore \frac{\Delta p_1}{p_1^0} & =x_2 \\ \therefore \frac{p_1^0-p_1}{p_1^0} & =x_2 \\ \therefore \frac{p_1^0-p_1}{p_1^0} & =\frac{n_2}{n_1+n_2} \quad\left(\because x_2=\frac{n_2}{n_1+n_2}\right)\end{aligned}$ Where, $n_1=$ Moles of solvent $n_2=$ Moles of solute → For dilute solutions $n _2 \ll n _1$ $\begin{aligned} \frac{p_1^0-p_1}{p_1^0} & =\frac{n_2}{n_1} \\ \therefore \frac{p_1^0-p_1}{p_1^0} & =\frac{ W _2 \times M _1}{ M _2 \times W _1}\end{aligned}$ Where, $\begin{aligned} W _1 & =\text { Weight of solvent } \\ W _2 & =\text { Weight of solute } \\ M _1 & =\text { Molar mass of solvent } \\ M _2 & =\text { Molar mass of solute } \\ p_1^0 & =\text { Vapour pressure of pure solvent } \\ p_1 & =\text { Vapour pressure of solution }\end{aligned}$
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