Answer to Question #310633 in Chemistry for Ronabe Agluba

Question

What is the molality of 40g KBr dissolved 800 mL water

Answer 1

Solution:

solute = KBr

solvent = water (H₂O)

The density of water (H₂O) is around approximately 1 g/mL

Density = Mass / Volume

Mass = Density × Volume

Therefore,

Mass of H₂O = (1 g/mL) × (800 mL) = 800 g

Kilograms of H₂O = (800 g H₂O) × (1 kg / 1000 g) = 0.8 kg H₂O

The molar mass of KBr is 119 g/mol

Therefore,

Moles of KBr = (40 g KBr) × (1 mol KBr / 119 g KBr) = 0.336 mol KBr

Molality = Moles of solute / Kilograms of solvent

Therefore,

Molality of KBr solution = Moles of KBr / Kilograms of H₂O

Molality of KBr solution = (0.336 mol) / (0.8 kg) = 0.42 mol/kg = 0.42 m

Molality of KBr solution = 0.42 m

Answer: The molality of KBr solution is 0.42 m