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24 December, 10:04

Given stock solutions of glucose (1M), asparagine (100 mM), and NaH2PO4 (50 mM), how much of each solution would you need to prepare 500 mL of a reagent that contains 0.05 M glucose, 10 mM asparagine, and 2 mM NaH2PO4?

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  1. 24 December, 10:13
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    25 mL, 50 mL, 20 mL

    Explanation:

    Molarity = numbers of mole / volume in liters

    Glucose

    stock solutions concentration = 1M

    concentration needed = 0.05 M

    volume of the preparation = 500ml = 500 / 1000 = 0.5l

    number of moles of glucose in the solution = Molarity * molar mass = 0.05 * 0.5 L = 0.025 moles

    volume of the stock needed = number of moles / molarity of the stock solution = 0.025 moles / 1 M = 0.025L = 25 mL

    Asparagine

    Molarity of the stock solution = 100mM = 0.1 M

    Molarity needed = 10mM = 0.01 M

    volume of the prepared solution = 0.5 L

    number of moles in the prepared solution = 0.01 M * 0.5 L = 0.005 moles

    volume of the stock solution needed = number of moles of the prepared solution / Molarity of the stock = 0.005 / 0.1 = 0.05L = 50 mL

    NaH₂PO₄

    Molarity of the stock solution = 50 mM = 0.05 M

    Molarity of the needed = 2 mM = 0.002 M

    Volume of the prepared solution = 0.5 L

    number of moles in the prepared solution = 0.002 M * 0.5 L = 0.002 * 0.5 L = 0.001 moles

    volume of the stock needed = 0.001 moles / 0.05 M = 0.02 L = 20 mL
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