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How much of glucose (C6H1206) is needed to make 1 L of a 1-M solution? Use details to support your answer.

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Answer:

200 g C₆H₁₂O₆

General Formulas and Concepts:

Chemistry - Solutions

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Molarity = moles of solute / liters of solution

Explanation:

Step 1: Define

1 M C₆H₁₂O₆

1 L of solution

Step 2: Identify Conversions

Molar Mass of C - 12.01 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

Step 3: Find moles of solute

1 M C₆H₁₂O₆ = x mol C₆H₁₂O₆ / 1 L

x = 1 mol C₆H₁₂O₆

Step 4: Convert

[tex]1 \ mol \ C_6H_{12}O_6(\frac{180.18 \ g \ C_6H_{12}O_6}{1 \ mol \ C_6H_{12}O_6} )[/tex] = 180.18 g C₆H₁₂O₆

Step 5: Check

We are given 1 sig figs. Follow sig fig rules and round.

180.18 g C₆H₁₂O₆ ≈ 200 g C₆H₁₂O₆

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