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How many grams of K2SO4 are required to prepare 750 mL of a 0.050 M solution?

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How many grams of K2SO4 are required to prepare 750 mL of a 0.050 M solution?

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How many grams of K2SO4 are required to prepare 750 mL of a 0.050 M solution?

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Answer

  • Mass = 6.53 grams
    Explanation:
    How many grams of K2SO4 are required to prepare 750 mL of a 0.050 M solution?
    Concentration = Moles / Volume (L)
    Moles = mass / molar mass
    Moles = Concentration * Volume (L)
    Moles = 0.050 M * 0.750 L
    = 0.0375 moles
    moles = mass / molar mass
    0.0375 = mass / 174.259 g/mol
    Mass = 174.259 * 0.0375
    Mass = 6.53 grams

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Answered on June 21, 2020 8:14 am
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