An aqueous solution of methanol and water has vapour pressurea)Equal to that of waterb)Equal to that of methanolc)More than that of waterd)Less than that of waterCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Yes it is option c...as methanol is there we can tell it as positive deviation...and in positive deviation vapour pressure of mixture ( methanol + h2o)... is greater than that of vapour pressure of solution ( here it is h2o)

Answer

The aqueous solution of methanol and water doesn't follow the Raoult's law over the entire range of concentration. And the solution of them shows the positive deviation from the Raoult's law. As some of the hydrogen bonds of the host ( here water) gets break down. Due to which vapour pressure of solution becomes more than that of water.
In this case the interaction between A-A and B-B is stronger than that of between A-B.

Answer

Explanation:

Composition of the solution:
- The solution is a mixture of methanol and water.

Vapour pressure of the solution:
- The vapour pressure of a solution is determined by the partial pressures of the components in the gas phase.
- Methanol has a lower boiling point compared to water, so it evaporates faster.
- As methanol evaporates, it increases the partial pressure of the vapour in the solution.
- This results in the overall vapour pressure of the solution being higher than that of pure water.

Key Points:
- The vapour pressure of the solution is more than that of water.
- This is because methanol has a lower boiling point and evaporates faster, increasing the overall vapour pressure of the solution.

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The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.?

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