Question

A solution is made by dissolving 18.29g of glucose C₆H₁₂O₂ in 50mL of water at 25C Calculate the molality,molarity, percent mass, Assume density of water is 1.00g/ml

Answer 1

density of water is 1.00g/ml then 50 mL = 50 g = 0.05 kg

Molarity (M) = moles / liter = mol/L

molality (m) = moles/ per kilogram of solvent (water) = mol/kg

percent mass = [mass of solute / mass of solution] x 100%

mass of solution = mass of solute + mass solvent = 18.29 g + 50 g = 68.29 g

moles = mass of solute (glucose) in grams / molar mass of solute (glucose) in grams/mol)]

moles of solute = 18.29 g / 180.156 g/mol = 0.1015 mol

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Molarity calculations:

Molarity = moles x (1000 / volume in mL) = 0.1015 mol x (1000 / 50) = 2.03 M

Molarity = 2.03 M

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molality calculations:

mass of 50 mL water in kg = 50/1000 = 0.05 kg

molality = 0.1015 moles / 0.05 kg = 2.03 m

molality = 2.03 m

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percent mass calculations:

percent mass = (18.29 g / 68.29 g) x 100% = 26.78 %

percent mass = 26.78%

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