What is the molality of a water solution that boils at 101.70°C?

Question

Wyatt Adkins · Accepted Answer

The molality of the solution is 3.32 mol/kg.

Water has a #K_b# of #"0.512 °C·kg·mol"^"-1"# of ions or molecules in solution. This indicates that the interference with the water molecules will raise the boiling point by 0.512 °C for each mole of particles in 1 kg of water. Water normally boils at 100.00 °C. 101.70 - 100.00 = 1.70 indicates a 1.70 °C increase in boiling point. #ΔT_b = K_bm# ∴ #m =( ΔT_b)/K_b = "1.70 °C"/("0.512 °C·kg·mol"^"-1") = "3.32 mol·kg"^"-1"# There is a 3.32 m molality. This molality could be achieved by dissolving 1 mol of a salt like #"CaCl"_2#, which ionizes into three particles, in 1 kg of water or by dissolving 3 mol of a compound like sugar in 1 kg of water.

Henry Antrim · Answer

undefined The molality of a water solution that boils at 101.70°C would depend on the concentration of solute dissolved in the water. Without knowing the concentration of the solute, it's not possible to determine the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. Therefore, the molality can only be calculated if the amount of solute and the mass of the solvent (in this case, water) are known.