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28 December, 05:57

How many grams of lithium fluoride is required to make 1.2 L of a 3.5 M

solution?

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Answers (1)
  1. 28 December, 06:18
    0
    109.2g

    Explanation:

    First, let us obtain the number of mole of lithium fluoride in the solution. This is illustrated below:

    Molarity of lithium fluoride (LiF) = 3.5M

    Volume = 1.2L

    Number of mole of LiF = ?

    Molarity = mole/Volume

    Mole = Molarity x Volume

    Number of mole LiF = 3.5 x 1.2

    Number of mole LiF = 4.2 moles

    Now let us convert 4.2 moles to grams in order to obtain the desired result. This illustrated below:

    Molar Mass of LiF = 7 + 19 = 26g/mol

    Number of mole of LiF = 4.2 moles

    Mass of LiF = ?

    Mass = number of mole x molar Mass

    Mass of LiF = 4.2 x 26

    Mass of LiF = 109.2g

    Therefore, 109.2g of lithium fluoride is required.
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