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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?
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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