Calculate the number of moles of solute present in each of the following aqueous solutions: (c) 124.0 g of a solution that is 6.45% glucose (C₆H₁₂O₆) by mass.

검증된 단계별 안내

Identify the mass percentage of glucose in the solution, which is given as 6.45%.

Calculate the mass of glucose in the solution by multiplying the total mass of the solution (124.0 g) by the mass percentage (6.45%) expressed as a decimal.

Determine the molar mass of glucose (C_6H_12O_6) by adding the atomic masses of all the atoms in the molecular formula: 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.

Use the mass of glucose obtained in step 2 and the molar mass from step 3 to calculate the number of moles of glucose using the formula: \( \text{moles} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} \).

The result from step 4 will give you the number of moles of glucose present in the solution.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Percent by Mass

Percent by mass is a way to express the concentration of a solution, indicating the mass of solute present in a given mass of solution. In this case, a 6.45% glucose solution means that there are 6.45 grams of glucose for every 100 grams of solution. This concept is essential for determining the mass of solute in a specific mass of solution.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For glucose (C6H12O6), the molar mass is approximately 180.18 g/mol. Knowing the molar mass allows us to convert between grams of solute and moles, which is crucial for calculating the number of moles in a solution.

Moles of Solute

The number of moles of solute can be calculated using the formula: moles = mass (g) / molar mass (g/mol). This relationship is fundamental in stoichiometry and solution chemistry, as it allows us to quantify the amount of solute present in a solution, which is necessary for various calculations in chemistry.

Calculate the number of moles of solute present in each of the following aqueous solutions: (c) 124.0 g of a solution that is 6.45% glucose (C6H12O6) by mass.