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प्रश्न
State Raoult law and obtain the expression for lowering of vapour pressure when the nonvolatile solute is dissolved Insolvent.
उत्तर
In an ideal solution, the vapour pressure of the solution is decreased when a non-volatile solute is dissolved in a solvent. The magnitude of decrease in the vapour pressure of the solution depends on the amount of solute added.
Let us consider the solution with the following features.
Mole fraction of the solvent = xA
Mole fraction of the solute = xB
Vapour pressure of the pure solvent = P°A
Vapour pressure of solution = P
As the solute is nonvolatile, the vapour pressure of the solution is only due to the solvent.
Therefore, the vapour pressure of the solution (P) will be equal to the vapour pressure of the solvent (PA) over the solution.
i.e., P = PA
According to Raoult’s law, the vapour pressure of solvent over the solution is equal to the product or its vapour pressure in a pure state and its mole fraction.
PA = P°A xA or
P = P°A xA
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संबंधित प्रश्न
Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot `"P"_"total"`, `"P"_"chloroform"` and `"P"_"acetone"` as a function of `"x"_"acetone"`. The experimental data observed for different compositions of mixtures is:
| `bb(100 xx "x"_"acetone")` | 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
| `bb("P"_"acetone"//"mm Hg")` | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| `bb("P"_"chloroform"//"mm Hg")` | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
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