1. How many mg of lithium phosphide are in 18.0 mL of a 1.25 M solution?

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1. How many mg of lithium phosphide are in 18.0 mL of a 1.25 M solution?

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Orla Orla 4 years 2021-08-27T19:31:10+00:00 1 Answers 7 views 0

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  1. Eirian
    0
    2021-08-27T19:32:39+00:00
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    Answer:

    1160mg

    Explanation:

    Molarity = number of moles ÷ volume

    According to the information in the question, molarity = 1.25 M, volume = 18.0 mL = 18/1000 = 0.018L

    M = n/V

    n = M × V

    n = 1.25 × 0.018

    n = 0.0225moles.

    Using mole = mass/molar mass, to find the mass of lithium phosphide (Li3P)

    Molar mass of Li3P = 6.9(3) + 31 = 51.7g/mol

    mole = mass/molar mass

    0.0225 = mass/51.7

    mass = 1.16grams.

    In milligrams (mg), mass of Li3P = 1.16 × 1000 = 1160mg

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