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How many grams of Magnesium Chloride (molar mass = 95.2 g/mole) will be needed to make 6.0 L of a 2.5 M MgCl2 solution?

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SvetkaChem

Answer:

1428 g

Explanation:

2.5 M = 2.5 mol/L

2.5mol/L*6.0 L = 15 mol MgCl2.

Molar mass (MgCl2) = 24.3 g/mol+(2*35.5 g/mol)=95.2 g/mol

15 mol * 95.2 g/mol = 1428 g

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