Question

How much water should be added to 4.3 moles of LiBr to prepare a 2.05 m solution?

Answer 1

Answer:

2.1 kg of water

Explanation:

Step 1: Given data

• Moles of lithium bromide (solute): 4.3 moles

• Molality of the solution (m): 2.05 m (2.05 mol/kg)

• Mass of water (solvent): ?

Step 2: Calculate the mass of water required

Molality is equal to the moles of solute divided by the kilograms of solvent.

m = moles of solute/kilograms of solvent

kilograms of solvent = moles of solute/m

kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg