Question

A solution of Sulphuric Acid in water exhibits:
Option A: Negative deviations from Raoult’s law.
Option B: Positive deviations from Raoult’s law.
Option C: Ideal properties.
Option D: The applicability of Henry’s law.

Answer 1

Hint: If we are aware of the concept of Raoult’s law, we will be able to crack the code for the problem. Water is a polar solution and Sulphuric acid is soluble clearly in water. Now if you apply the law and the points mentioned, you will be able to derive the answer.
Law used:
Raoult’s Law: It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture. Hence, the relative lowering of vapour pressure of a dilute solution of nonvolatile solute is equal to the mole fraction of solute in the solution.

Complete solution:
Explanation of terms used in this question :
Partial pressure: It is the pressure which is exerted by an individual gas/liquid in a mixture of gases/liquids.
Ideal mixture: It is the mixture where any part of the mixture is same in concentration, temperature and pressure.
Vapour Pressure: It is the measure of ability of a substance to get converted into gaseous state or vapour state.
Pure component: It is the ideal component which stays in the same state when the state of mixture is changed.
Mole Fractions: It is the ratio of moles of a particular component to the total number of moles of a mixture.
When we take water and add some amount of Sulphuric Acid , the Sulphuric acid readily dissolves in the solution among the polar molecules of water. Since the dissolving is done in a quick and clear manner, a stronger interaction is now formed between the newly formed ions (ions of Sulphuric acid and water) than the interaction between the Sulphuric acid ions.
Since the interactions formed are way too strong, extra heat is needed to make the molecules to vaporize. Hence, a considerable decrement takes place in the vapour pressure of the solution. Change in vapour pressure due to change in the strength of interaction is a deviation in Raoult’s law. Since the vapour pressure in this state decreases, it is said to be a negative deviation, which is the second option.
Hence the answer is Option A- Negative deviations from Raoult’s law.

Note: The question need not be asked in the same way as of this type. The solvent may vary. That is, they can give a nonpolar solution instead of water like Benzene and Carbon Tetrachloride. In such cases, if the interactions are weaker, then it means the vapour pressure increases and the behavior exhibits positive deviation.