This is general chemistry

Question

A solution is prepared by dissolving 318.6 g sucrose (C12H22O11) in 4905 g of water. Determine the molarity of the solution

Answer 1

I'm guessing that maybe the question actually asked to calculate the molality of the solution since they gave you grams of water and not mls of water. Also, they didn't provide you with the density of water, and finally, if you added 318 g of sucrose to 4905 ml of water, the final volume would NOT be 4905 ml but would certainly be greater than that. If that's the case, then the following is the procedure. If not, then the previous answer is the only way to attempt to answer this, even though one really doesn't know the final volume.

molality = moles solute (sucrose) per kg of solvent (water)

moles sucrose = 318.6 g/342 g = 0.9316 moles

kg solvent = 4905 g x 1 kg/1000 g = 4.905 kg

molality = 0.9316 mol/4.905 kg = 0.1899 m

Answer 2

Molarity = mols of solute / liters of solution

Convert 318.6 g of sucrose to mols

318.6 g / 342 g = 0.9316 mols

The density of water is 1 g / mL. So if it's 100 g of water, it's 100 mL.

In this case, 4905 g of water is 4905 mL. Convert this to liters and it's 4.905 L

0.9316 mols / 4.905 L = 0.1899 M

Answer 3

molarity = moles of solute / liters of solution

318g sucrose 1mol sucrose 1000 mL

——————— x —————————— x —————— = 0.189 mol / L

4905 ml H₂O 342g C₁₂H₂₂O_{11 1 Liter}