Question

A solution of two miscible liquids showing negative deviation from Raoult's law will have:

• A. Increased vapour pressure, increased boiling point
• B. Increased vapour pressure, decreased boiling point
• C. Decreased vapour pressure, decreased boiling point
• D. Decreased vapour pressure, increased boiling point

Answer 1

The Correct Option is D

To understand the behavior of a solution of two miscible liquids showing negative deviation from Raoult's Law, let's explore the concepts involved:

Raoult's Law 

According to Raoult's Law, the partial vapor pressure of each component in an ideal solution is directly proportional to its mole fraction. Mathematically, it is expressed as:

\(P_A = x_A \cdot P^0_A\)

Where:

• \(P_A\) is the partial vapor pressure of component A.
• \(x_A\) is the mole fraction of component A in the solution.
• \(P^0_A\) is the vapor pressure of pure component A.

Negative Deviation

A negative deviation from Raoult's Law occurs when the interactions between different components of the solution are stronger than those in the pure liquids. This leads to a situation where:

• The vapor pressure of the solution is lower than expected from Raoult’s Law.
• The solution shows exothermic mixing (release of heat).
• Stronger adhesive forces lead to decreased vapor escapement, thus reducing vapor pressure.

Boiling Point

The boiling point of a solution is the temperature at which its vapor pressure equals the external pressure. Since the vapor pressure is decreased, more heat is required to reach the boiling point compared to the pure solvents.

Conclusion

Thus, for a solution showing negative deviation from Raoult's Law:

• The vapor pressure is decreased.
• The boiling point is increased, as more heat is necessary to reach the vapor pressure of boiling.

Therefore, the correct answer is: Decreased vapour pressure, increased boiling point.

Answer 2

A solution showing negative deviation from Raoult’s law has lower vapour pressure than expected because the interactions between the components are stronger than in the pure liquids. This results in a decreased vapour pressure, which leads to an increase in boiling point.

For solutions showing negative deviation:

$$ P_T < P_A^0 X_A + P_B^0 X_B $$

where \( P_T \) is the total vapour pressure, \( P_A^0 \) and \( P_B^0 \) are the vapour pressures of pure components, and \( X_A \) and \( X_B \) are their mole fractions.