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27 November, 03:25

How many milliliters of a 0.205 M solution of glucose, C6H12O6, are required to obtain 150.1 g of glucose?

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  1. 27 November, 07:02
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    4050 mL

    Explanation:

    Given dа ta:

    Mass of glucose = 150.1 g

    Molarity of solution = 0.205 M

    Volume of solution = ?

    Solution:

    Molarity = number of moles of solute / L of solution.

    Now we will calculate the moles of sugar first.

    Number of moles = mass / molar mass

    Number of moles = 150.1 g / 180.156 g/mol

    Number of moles = 0.83 mol

    Now we will determine the volume:

    Molarity = number of moles of solute / L of solution.

    0.205 M = 0.83 mol / L of solution.

    L of solution = 0.83 mol / 0.205 M

    L of solution = 4.05 L

    L to mL conversion:

    4.05 L * 1000 mL / 1 L = 4050 mL
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