Answer:
1) 71.9 g/L
2) 0.221 m
olal
3) 7.05% by mass
Explanation:
Step 1: Data given
Concentration of sucrose = 0.210 M
Molar weight of sucrose = 342.3 g/mol
Density of solution = 1.02 g/mL
Mass of water = 948.1 grams
Step 2: Convert this value into terms of g/L
(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L
Calculate the molality
Step 1: Calculate mass water
Suppose we have a volume of 1.00L
Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution
We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means
(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water
Step 2: Calculate molality
Molality = moles sucrose / mass water
(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m
olal
Mass %
% MAss = (mass solute / mass solution)*100%
(71.9 g) / (1020 g) *100% = 7.05% by mass
