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