Calculate the molarity of 101 grams of sucrose, C12H22O11, in 2.50 liters of solution.

Answer 1

In order to answer this question, you need to know the units for a molar solution (M). M = mol/L

So to solve this problem you need to:

1. Find the number of moles of sucrose in 101 grams by dividing 101 by the molar mass of sucrose and then
2. 101 g sucrose x (1 mol sucrose/ 342.30 g) = 0.295 mol
3. Divide the number of moles by the liters of solution
4. 0.295 mol / 2.5 L = 0.118 M <----- that's your answer!

Answer 2

Molar mass C₁₂H₂₂O₁₁ = 342 g/mole

molarity = moles/liter

moles = 101 g x 1 mole/342 g = 0.295 moles

liters = 2.50

molarity = 0.295 moles/2.50 liters = 0.118 M