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22 November, 23:27

How many grams of CaBr2 would be needed to create 450. mL of a 2.00 M

solution? Round off your answer to the nearest gram. Record your answer in the

grid below.

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Answers (1)
  1. 22 November, 23:30
    0
    Mass of CaBr₂ = 179.9 g

    Explanation:

    Given dа ta:

    Volume of CaBr₂ solution = 450.0 mL (450/1000 = 0.45 L)

    Molarity of solution = 2.00 M

    Mass of CaBr₂ = ?

    Solution:

    Formula:

    Molarity = number of moles / volume of solution in L

    Now we will put the values.

    2.00 M = number of moles / 0.45 L

    Number of moles = 2 M * 0.45 L

    Number of moles = 0.9 mol (M = mol/L)

    Mass of CaBr₂:

    Mass of CaBr₂ = number of moles * molar mass

    Mass of CaBr₂ = 0.9 mol * 199.89 g/mol

    Mass of CaBr₂ = 179.9 g
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