Respuesta :
Answer:
Compound B has greater molar mass.
Explanation:
The depression in freezing point is given by ;
[tex]\Delta T_f=i\times k_f\times m[/tex]..[1]
[tex]m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}[/tex]
Where:
i = van't Hoff factor
[tex]k_f[/tex] = Molal depression constant
m = molality of the solution
According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.
The depression in freezing point of solution with A solute: [tex]\Delta T_{f,A}[/tex]
Molar mass of A = [tex]M_A[/tex]
The depression in freezing point of solution with B solute: [tex]\Delta T_{f,B}[/tex]
Molar mass of B = [tex]M_B[/tex]
[tex]\Delta T_{f,A}>\Delta T_{f,B}[/tex]
As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.
[tex]\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}[/tex]
[tex]M_A<M_B[/tex]
This means compound B has greater molar mass than compound A,
