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?

Expert's answer

"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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