Question

A student caries out an experiment and dissolves 10.0 g of sodium bromide. NaBr, in 200g of water. What is the molality of the solution?

Answer 1

0.485mol/kg

Step-by-step explanation:

Molality is defined as the ratio of the number of moles of a solute to the kilogram of a solvent which is usually water.

Mass of NaBr = 10g

Molar mass of NaBr= 23+80=103g/mol

Mole of NaBr = mass / molar mass=10g/103g/mol= 0.097mol

Molality= Mole of a solute/ mass of solvent in kg

Mole of a solute( NaBr)= 0.097mol; mass of solvent( water in kg)= 200g. To convert to kg: 200g/1000g= 0.2kg

Molality = 0.097mol/0.2kg = 0.485mol/kg