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Answer to Question #311731 in Inorganic Chemistry for Careng

Question #311731

What weight of magnesium chloride is required to prepare 202 ml of 0.105M



solution?

1
Expert's answer
2022-03-15T21:38:01-0400

"m(MgCl_2)=202\\ mL\u00d7\\frac{0.105\\ mol}{1000\\ mL}\u00d7\\frac{95.22\\ g}{1\\ mol}=2.02\\ g"


Answer: 2.02 g


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