A solution was made where 24.0 grams of glucose C6H12O6, were dissolved in 1.00 kg of water, the density of this solution was 1.07 g/mL

Question

Find the molality of this sugar solution

Find the mole fraction of the glucose

Find the mole fraction of the water

Find the mass % of glucose in this sugar solution

Find the molality of this solution

Answer 1

The question asks for molality twice, so I'll assume the last question was meant to ask for molarity.

molar mass glucose = 180 g/mole

molality = moles glucose/kg water: moles = 24.0 g/180 g/mol = 0.133 mol/1.00 kg = 0.133 m

mole fraction glucose = moles glucose/total moles: moles glucose = 0.133. Moles water = 1000 g/18 g/mol = 55.6 moles. Total moles = 0.133 + 55.6 = 55.7. Mole fraction = 0.133 mol/55.7 mol = 0.00239

mole fraction water = moles water/total moles = 55.6/55.7 = 0.998

mass % glucose = mass glucose/total mass (x100%) = 24.0g/1024 g(x100%) = 2.34%

molarity = moles glucose/liter solution: moles = 0.133. Volume of solution = 1024 g x 1 ml/1.07 g = 957 ml = 0.957 liters. molarity = 0.133 mol/0.957 L = 0.139 M