Question

A chemist prepares two solutions: one from 28 g potassium bromide (KBr) and 1 kg water; and the other from 28 g sodium iodide (NaI) and 1 kg water. What information is most useful to determine the solution that has the lower freezing point? the density of each solution in grams per liter the crystal structures of KBr and NaI the molar masses of KBr and NaI the molar mass of water

Answer 1

Answer: Molar masses of KBr and NaI

Explanation:

$\Delta T_b=K_b\times molality=K_b\times \frac{\text{Mass of the solute}}{\text{Molar mass of the solute}\times \text{Weight of solvent in kg}}$

From the the above formula ,we can say that $\Delta T_b$ is inversely related to molar mass of the compound. Lower the value of molar mass more will be the value of $\Delta T_b$ which means more will be the lowering in the freezing temperature of that solution and vice-versa.

$\Delta T_b\propto \frac{1}{\text{Molar mass of the compound}}$

For KBr

$K_b$ for water is = 0.59 K kg/mol

$\Delta T_b=0.59 K kg/mol\times \frac{28g}{119g/mol\times 1kg}=0.1388 K$

For NaI

$K'_b$ for water is = 0.59 K kg/mol

$\Delta T'_b=0.59 K kg/mol\times \frac{28g}{149.89g/mol\times 1kg}=0.1102 K$

$\Delta T_b>\Delta T'_b$

Since, all the values beside molar mass are same. The information of molar masses of KBr and NaI will be most useful to determine the solution that has the lower freezing point.