Maharashtra BoardEnglish MediumSTD 12 ScienceChemistrySolutions4 Marks
Question
State and explain Raoult’s law.
✓
Answer
Statement of Raoult’s law : The law states that, at constant temperature, the partial vapour pressure of any volatile component of a solution is equal to the product of vapour pressure of the pure component and the mole fraction of that component in the solution. Let $P _0$ and $P$ be the respective vapour pressures of a pure volatile component and a solution. If $x _1$ is the mole fraction of a solvent then by Raoult's law, $P=x_1 \times P_0$
Explanation : Consider a solution containing two volatile components $A$ and $B$ having mole fractions $x_1$ and $x_2$ respectively. Let $P_1^0$ and $P_2^0$ be the vapour pressures of pure components (or liquids) $A$ and $B$ respectively. Then by Raoult's law, vapour pressure of component $A = P _1= X _1 \times P_1^0$, vapour pressure of component $B=P_2=x_2 \times P_2^0$ Here $P_1$ and $P_2$ represent partial vapour pressures of the two liquid components in the solution. Hence the total vapour pressure, $P_T$ of the solution will be,
$\begin{aligned} & P _{ T }= P _1+ P _2 \\ & \therefore P _{ T }= x _1 P_1^0+ x _2 P_2^0 \\ & \because x _1+ x _2=1 \\ & \therefore x _1=1- x _2 \\ & \therefore P _{ T }=\left(1- x _2\right) P_1^0+ x _2 P_2^0 \\ & =P_1^0- x _2 P_1^0+ x _2 P_2^0 \\ & =\left(P_2^0-P_1^0\right) x _2+P_1^0 \end{aligned}$
With the help of above equation, the vapour pressures of solutions having different concentrations (or mole fractions) can be calculated.
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