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------------------------------------------------ What is the molarity of an aqueous solution containing 22.5 g of sucrose (C₁₂H₂₂O₁₁) in 35.5 mL of solution?

The molarity of an aqueous solution containing 22.5 g of sucrose in 35.5 mL of solution is calculated by first determining the number of moles of sucrose in the solution, then converting the volume to liters, and finally dividing the moles by the volume in liters. The calculated molarity is 1.85 M.

To calculate the molarity of the sucrose solution, we need to follow a set of steps. Firstly, we would need to calculate the number of moles of sucrose in the given mass. The molecular weight of sucrose (C₁₂H₂₂O₁₁) is 342.3 g/mol. Therefore, the number of moles of sucrose is 22.5 g / 342.3 g/mol = 0.0657 mol.

Then, convert the volume of the solution from milliliters to liters, as the molarity is defined as moles per liter. 35.5 mL is equal to 0.0355 L.

The molarity is then calculated by dividing the number of moles of solute by liters of solution. So here Molarity = 0.0657 mol / 0.0355 L = 1.85 M.

The sucrose in the solution remains dispersed, even though sucrose molecules are heavier than water molecules (according to the definition of an aqueous solution). Hence, the molarity of the sucrose solution is 1.85 M.

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