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The practical E1 requires 1000 mg Zn/L. Calculate the mass of Zn (NO3)2·6H2O needed to prepare...

The practical E1 requires 1000 mg Zn/L. Calculate the mass of Zn (NO3)2·6H2O needed to prepare 500 ml of the stock solution 1000 mg Zn/L. Express the results in grams and using 3 digits.

MW Zn (NO3)2·6H2O: 297.49g

MW Zn: 65.39g

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Homework Answers

Answer #1

1000 mg Zn / L mean 1000 mg Zn for 1000 ml solution then for 500 ml solution required Zn = 500 mg

Molar mas of Zn (NO3)2·6H2O = 297.49 gm / mole

molar mass of Zn = 65.39 gm /mole

1 mole of Zn (NO3)2·6H2O contain 1 mole of Zn that mean 297.49 gm of Zn (NO3)2·6H2O contain 65.39 gm of Zn then for 500 mg Zn required Zn (NO3)2·6H2O = 500 297.49 / 65.39 = 2274.74 mg of Zn (NO3)2·6H2O

2274.74 mg of Zn (NO3)2·6H2O requred to prepare 500 ml of 1000 mg Zn/L stock solution

2.275 gm of Zn (NO3)2·6H2O requred to prepare 500 ml of 1000 mg Zn/L stock solution

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