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Solutions — Chemistry STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceChemistrySolutions3 Marks
Question
Explain the variation of vapour pressure with mole fraction of a solvent in solution.
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Explain the variation of vapour pressure with the concentration of a solution.
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Answer

The vapour pressure of a pure solvent decreases when a nonvolatile solute is dissolved in it. Consider a pure solvent with vapour pressure $P_0$ and mole fraction $x _1$. For a solution containing a non-volatile solute, if $x _1$ and $x_2$ are the mole fractions of a solvent and a solute respectively, then $x_1+x_2=1$ and $x_1<1$.
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By Raoult's law, $P = x _1 \times P _0$. As the mole fraction of a solvent in the solution increases the vapour pressure increases as shown in the above figure 2.5 . When $x _1$ becomes equal to 1 , the vapour pressure becomes $P _0$, i.e. the vapour pressure of a pure solvent.

If at any mole fraction of a solvent, the vapour of the solution is $P$, then the lowering of vapour pressure will be, $\Delta P=P_0-P$.

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