How many grams of CuSO4*5H20 are needed to prepare a 0.020 M stock solution?

Question

I'm in General Chemistry in college and just need the answer to this question. It'd be really cool if you could show/explain how to do this if you can. That would be a huge help. Thanks for your time.

Answer 1

Megan,

 

Michael P has given the right answer but you might be struggling with it a bit so I will break it down further for you.

 

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A 0.02 M solution implies it has 0.02 mole of that substance (solute)  in 1 L of water (solvent).

 

1 mole of CuSO4.5H20 equals 249.685 g

 

Therefore, 0.02 mole of that substance equals 4.994 g

 

By adding 4.994 g of CuSO4.5H20 to 1 L of water you will get a 0.02 M solution

Answer 2

Megan,

 

You need the molecular weight of CuSO₄*5H₂O: 1Cu + 1S + 4O + 10H + 5O = __ grams of CuSO4*5H2O / 1 mole of CuSO4*5H2O. I'll let you add up the molecular weights of the five elements.

 

A 0.020M (molar) stock solution has 0.020 moles of CuSo4*5H2O in it with one liter of water.

 

(0.020 moles CuSO4*5H2O) * ( __ grams of CuSO4*5H2O/ 1 mole of CuSO4*5H2O ) = __ grams of CuSo4*5H2O.

 

Substitute the molecular weight that you calculated in the first line and multiply.

 

So,  grams of grams of CuSo4*5H2O is the same as 0.020 moles of grams of CuSO4*5H2O, and that is what you put in your one-liter volumetric flask before you add the water.

 

Please excuse me if I don't do the arithmetic. Thanks.

Michael.