Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

Solution problem check/help?

1. How many grams of MgCl2 are required to prepare 200 ml of a 0.35 M solution (MW of MgCl2 = 95.2)? 6.66 g

2. Express the concentration of the solution in problem #1 in terms of g/liter. 33.3 g/L

3. Express the concentration of the solution in problem #1 in terms of %w/v. 

1 Answer

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  • Fern
    Lv 7
    1 month ago
    Favorite Answer

    0.35 moles MgCl2/liter x 200 mL x 1 liter/1000 mL = 0.070 moles MgCl2

    0.070 moles MgCl2 x 95.2 grams MgCl2/mole MgCl2 - 6.7 grams MgCl2

    6.7 grams MgCl2/200 mL x 1000 mL/liter = 33 grams MgCl2/liter

    6.7 grams MgCl2/200 mL x 100 = 3.3%

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