What is the molality of a solution of 0.50 g of glucose, C6H12O6, 14.3 g of water?
To calculate the molality of a solution, we first need to find the moles of the solute (glucose) and then divide it by the mass of the solvent (water) in kilograms.
1. Calculate the moles of glucose:
molar mass of glucose (C6H12O6) = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
moles of glucose = 0.50 g / 180.18 g/mol = 0.002774 mol
2. Calculate the molality:
mass of water = 14.3 g
mass of water in kg = 14.3 g / 1000 = 0.0143 kg
molality = moles of solute / mass of solvent in kg
molality = 0.002774 mol / 0.0143 kg = 0.1937 mol/kg
Therefore, the molality of the solution is 0.1937 mol/kg.