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22 September, 19:19

What mass (in grams) of MgBr2 would be present in 23.7 mL of 0.875 M MgBr2 solution? Give your answer to 2 decimal places. (You'll need your periodic table)

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  1. 22 September, 22:10
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    A) 0.875 M of MgBr2 is 0.875 mol/L

    B) 23.7 mL = 0.0237 L

    0.875 mol = > 1 L

    x mol = > 0.0237 L

    Cross multiply

    1x = 0.875 * 0.0237

    x = 0.0207 mol

    >> In 23.7 mL of 0.875 M MgBr2 solution there is 0.0207 moles of MgBr2

    C) Molar mass of MgBr2 (*) = 24.305 + (2 * 79.904) = 184.113 g/mol

    184.113 g = > 1 mol

    x g = > 0.0207 mol

    Cross multiply

    1x = 184.113 * 0.0207

    x = 3.8111 g

    >> 0.0207 moles of MgBr2 is equivalent to 3.81 g of MgBr2

    >> In 23.7 mL of 0.875 M MgBr2 solution there is 3.81 g of MgBr2

    (*) Use your periodic table
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