Answer:
Compound B.
Explanation:
The freezing point depression is a colligative property. It depends on the number of particles (moles) present in the solution.
[tex]\Delta T_{\text{f}} = K_{\text{f}} b[/tex]
where b is the molal concentration
[tex]b = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{m}{M}[/tex]
If m is constant (5 g), then
[tex]n \propto \dfrac{1}{M}[/tex]
The compound with the greater molar mass has fewer moles and therefore fewer particles to depress the freezing point.
That must be Compound B, because Compound A has the lower freezing point.
