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If you prepare 525 mL of a .650M MgCl2 solution, how many grams of solute would you need?
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If you prepare 525 mL of a .650M MgCl2 solution, how many grams of solute would you need?

User Sovalina
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Answer:

Step-by-step explanation:

525ml of 0.6550M MgCl₂ => 0.525L x 0.650M = 0.3413mole MgCl₂

0.3413mole MgCl₂ = 0.3413mole x 94 g/mole = 32.1g MgCl₂

User Nyxm
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