Question

A chemist dissolves 0.099 g of CuSO4 · 5 H2O in water and dilutes the solution...

A chemist dissolves 0.099 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 28.9-mL sample of this solution is then transferred to a second 500-mL volumetric flask and diluted. What is the molarity of CuSO4 in the second solution?

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Answer #1

Molar mass of CuSO4 5H2O = 249.685 g/mol.

So,

249.685 g of CuSO4 5H2O in 1000 mL of water = 1 M

0.099 g of CuSO4 5H2O in 1000 mL of water = (0.099/249.685) M

0.099 g of CuSO4 5H2O in 1000 mL of water = 3.965 x 10-4 M

0.099 g of CuSO4 5H2O in 500 mL of water = (3.965 x 10-4) / 2 M = 1.9825 x 10-4 M

Hence, in 500 mL of solution 1.9825 x 10-4 moles of CuSO4 5H2O present.

In 28.9 mL of solution (1.9825 x 10-4 x 28.9) / 500 moles of CuSO4 5H2O present.

In 28.9 mL of solution 1.146 x 10-5 moles of CuSO4 5H2O present.

In 28.9 mL of solution 2.86 x 10-3 gm of CuSO4 5H2O present.

Again,

249.685 gm of CuSO4 5H2O in 1000 mL = 1 M

249.685 gm of CuSO4 5H2O in 500 mL = 0.5 M

2.86 x 10-3 gm gm of CuSO4 5H2O in 500 mL = (0.5 x 2.86 x 10-3)/249.685 M = 5.727 x 10-3 M

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