How many moles of sucrose, C₁₂H₂₂O₁₁, is present in 80.0 grams of an aqueous solution that is 3.50% sucrose by mass?
MM(C₁₂H₂₂O₁₁) = 342.30 g/mol

Question

How many moles of sucrose, C₁₂H₂₂O₁₁, is present in 80.0 grams of an aqueous solution that is 3.50% sucrose by mass? MM(C₁₂H₂₂O₁₁) = 342.30 g/mol

Quizplus · Accepted Answer

The mass of sucrose in the solution is 3.50% of 80.0 grams, which is 2.8 grams. The number of moles of sucrose is calculated by dividing the mass of sucrose by its molar mass: 2.8 g / 342.30 g/mol = 8.18×10^-3 moles.

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How Many Moles of Sucrose, C₁₂H₂₂O₁₁, Is Present in 80