Final answer:
The molality of a 30.2 g sucrose solution in 285 g of water is 0.310 m. The molality of an 8.35 moles ethylene glycol solution in 1905 g of water is 4.38 m. Calculate the molality by dividing moles of solute by kilograms of solvent.
Step-by-step explanation:
To calculate the molality of a solution, you need to divide the number of moles of solute by the mass of the solvent in kilograms.
First, let's calculate the molar mass of sucrose which is C₁₂H₂₂O₁₁.
The atomic weights are as follows:
carbon (C) = 12.01 g/mol, hydrogen (H) = 1.008 g/mol, and oxygen (O) = 16.00 g/mol.
Add these together to get the molecular weight of sucrose:
(12 x 12.01) + (22 x 1.008) + (11 x 16.00) = 342.30 g/mol.
Now, for part (a), divide the mass of sucrose by its molar mass to get moles of sucrose:
30.2 g sucrose ÷ 342.30 g/mol = 0.0882 mol sucrose.
Convert the mass of water to kilograms:
285 g = 0.285 kg.
Finally, calculate molality:
0.0882 mol sucrose ÷ 0.285 kg water = 0.30984 mol/kg, or simply 0.310 m (to three significant figures).
For part (b), since we are already given the moles of ethylene glycol, we convert the mass of water to kilograms:
1905 g = 1.905 kg.
Then calculate the molality:
8.35 moles ethylene glycol ÷ 1.905 kg water = 4.38477 mol/kg, or 4.38 m (to three significant figures).