If 80 g of copper sulphate CuSO₄ . 5H₂O is dissolved in deionised water to make 5L of solution. The concentration of the copper sulphate solution is x × 10⁻³ mol L⁻¹. The value of x is _____________.
[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]

Solution

Chemistry ma’am · 3 steps

The 80 g is the blue hydrate, so divide by the molar mass that includes the five waters.

1. Molar mass of hydrate. The water of crystallisation is part of every gram you weigh. For CuSO₄.5H₂O: 63.54 + 32 + 4 × 16 + 5 × 18 = 249.54 g mol⁻¹.
2. Moles dissolved. n = 80/249.54 = 0.3206 mol. Each mole of hydrate gives one mole of CuSO₄ in solution.

   M = n/V = (0.3206 mol)/(5 L) = 0.0641 mol L⁻¹

3. So. 0.0641 mol L⁻¹ = 64.1 × 10⁻³ mol L⁻¹, so x is 64 to the nearest integer.

The common slip: 100. Using the anhydrous molar mass 159.54 g mol⁻¹ gives 0.1 mol L⁻¹, that is 100, not 64. The five waters were weighed too.

The idea to keep. When a salt is weighed as a hydrate, its molar mass must include the water of crystallisation. Moles of hydrate equal moles of the salt it releases.

Answer from the NTA final answer key. Solution by PrepWiser.