Question

On mixing $10\,mL$ of acetone with $40\,mL$ of chloroform, the volume of the solution would be which of the following?
A) $ < 50\,mL$
B) $ > 50\,mL$
C) $ = 50\,mL$
D) $Cannot \,be \,predicted$

Answer 1

Hint: Check for Hydrogen bonding between the molecules during the reaction. In chloroform, Carbon is attached to three Chlorine atoms and hence making it partially positive. This partially positive Carbon seeks electrons from Hydrogen transferring its partial positive charge to Hydrogen.

Complete answer:
Carbon in chloroform that is attached to three Chlorines becomes partially positive. This Carbon attracts electrons from the attached Hydrogen making it partially positive. Due to this partial positive charge, the Oxygen in acetone that is partially negative gets attracted to the Hydrogen creating weak Hydrogen bonding.
Due to the hydrogen bonding in our given molecules, Acetone-Chloroform bonds become stronger than Acetone-Acetone and Chloroform-Chloroform. This shows negative deviation from Roult’s Law which means that the total vapor pressure is lower than expected. This means that the volume of our mixture is lesser than we expect.

Therefore, the total volume of the mixture would be $ < 50\,mL$ i.e. option A.

Additional information:
According to Rault’s Law, the partial vapor pressure of a mixture is equal to the product of individual vapor pressure of the pure solvent and its mole fraction. Raoult's Law shows that as mole fraction of a solution decreases, its partial vapor pressure decreases as well.

Note:
When acetone and chloroform are mixed together, a hydrogen bond is formed between them which increase intermolecular interactions. Hence, Acetone-Chloroform interactions are stronger than Acetone-Acetone and Chloroform-Chloroform interactions.