Question

What is the molarity of a solution of sucrose, C₁₂H₂₂O₁₁, that contains 125 g of sucrose in 3.50 l of solution? (molar mass of C₁₂H₂₂O₁₁ = 342.34 g/mol)

Answer 1

To calculate the molarity of a sucrose solution, divide the moles of sucrose by the volume of the solution. With 125 g of sucrose and a solution volume of 3.50 L, the molarity is approximately 0.1043 M.

Step-by-step explanation:

To find the molarity of a sucrose solution, we can use the relationship molarity = moles of solute ∕ liters of solution. First, we need to calculate the number of moles of sucrose using the molar mass. The molar mass of sucrose, C₁₂H₂₂O₁₁, is given as 342.34 g/mol.

Moles of sucrose = mass of sucrose (g) ∕ molar mass (g/mol).

Moles of sucrose = 125 g ∕ 342.34 g/mol = 0.365 moles (approximately).

Now, we divide the moles of sucrose by the volume of the solution in liters:

Molarity (M) = 0.365 moles ∕ 3.50 L = 0.1043 M (approximately).

Thus, the molarity of the sucrose solution is approximately 0.1043 M.