To determine which pair will show positive deviation from Raoult's Law, we need to understand what Raoult's Law is and what positive deviation implies.
Raoult's Law states that the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent present in the solution. In a solution showing positive deviation from Raoult's Law, the observed vapor pressure is higher than predicted by Raoult's Law. This occurs when the interactions between unlike molecules are weaker than those among like molecules, leading to greater tendency for evaporation.
Let's analyze each pair:
Therefore, the correct answer is Benzene - Methanol, as the intermolecular forces between methanol and benzene are weaker than the forces within pure methanol, leading to positive deviation from Raoult's Law.
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \). 
Choose the correct answer from the options given below:
For x < 0:
f(x) = ex + ax
For x ≥ 0:
f(x) = b(x - 1)2