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(MgCl2)=202 mL×0.105 mol1000 mL×95.22 g1 mol=2.02 gm(MgCl_2)=202\ mL×\frac{0.105\ mol}{1000\ mL}×\frac{95.22\ g}{1\ mol}=2.02\ g


Answer: 2.02 g


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