# Calculate the mass of copper sulfate (CuSO4.5H2O) required to prepare 250 mL solution of 1000 mg Cu L-1.?

How do u do this type of question. Thanks

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Break the problem down into two parts, working backwards:

We need enough copper (as CuSO4.5H2O) to make 250 mL of a solution which contains 1000 mg of copper per litre. 250 mL is a quarter of a litre, so we need (1/4) x 1000 = 250 mg of copper.

The problem then is that copper sulphate contains other components beside copper, so how do we work out how much we need to get 250 mg of copper. If you are familiar with the concept of 'moles', it's reasonably easy.

The molar mass (or atomic weight) of copper is 63.546, so that 250 mg of copper is 250 x 10^-3/63,546 = 3.934 x 10^-3 moles. Since there is one atom of copper in each molecule of copper sulphate, this means that we need the same number of moles of that compound to ensure that we have 250 mg of copper.

The molar mass of copper sulphate is obtained by adding up the combined atomic weights of all the atoms present in the compound (1 x Cu, 1 x S, 9 x O and 10 x H) ie

Molar mass CuSO4.5H2O = 63.546 + 32.065 + 9 x 15.999 + 10 x 1.008 = 249.682 g

So 1 mole of copper sulphate weighs 249.682 grams; 3.934 x 10^-3 moles will weigh 3.934 x 10^-3 x 249.682 = 0.982 g.

Hence we need 0.982 g of copper sulphate to prepare 250 mL of the specified solution.

(If you're not familiar with moles, let me know and I'll rewrite it simply in terms of proportions.)

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