Calculate the solubility in mol $dm^{-3}$ of 40g of CuSO4 dissolved in 100g of water at 120°C? [Cu = 64, S = 32, O = 16]

1. Analyze the question

We are asked to find the solubility of Copper(II) tetraoxosulphate(VI) ($\text{CuSO}_4$) in $\text{mol dm}^{-3}$. We are given the mass of the solute (40g), the mass of the solvent (100g of water), and the atomic masses to calculate the molar mass.

2. Explain the concept

• Solubility in $\text{mol dm}^{-3}$: This is the molar concentration of a saturated solution. It is calculated as: $$\text{Solubility} = \frac{\text{Number of moles of solute}}{\text{Volume of solvent in dm}^3}$$
• Molar Mass: The sum of the atomic masses of all atoms in the formula.
• Density of Water: In these calculations, we assume the density of water is $1\text{g/cm}^3$ (or $1\text{g/ml}$). Therefore, $100\text{g}$ of water is equal to $100\text{cm}^3$, which is $0.1\text{dm}^3$.

3. Show the solution

Step 1: Calculate the molar mass of $\text{CuSO}_4$

• $\text{Cu} = 64$
• $\text{S} = 32$
• $\text{O} = 16 \times 4 = 64$
• $\text{Molar Mass} = 64 + 32 + 64 = 160\text{ g/mol}$

Step 2: Calculate the number of moles of $\text{CuSO}_4$

• $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$
• $\text{Moles} = \frac{40\text{g}}{160\text{g/mol}} = 0.25\text{ moles}$

Step 3: Convert the volume of water to $\text{dm}^3$

• $100\text{g of water} = 100\text{cm}^3$
• $100\text{cm}^3 = \frac{100}{1000}\text{dm}^3 = 0.1\text{dm}^3$

Step 4: Calculate the solubility

• $\text{Solubility} = \frac{0.25\text{ moles}}{0.1\text{dm}^3} = 2.5\text{ mol dm}^{-3}$

4. Identify the answer

Comparing our result to the options provided: A. 4.00 B. 2.50 C. 0.40 D. 0.25

The correct option is B.

5. Tip

When solving solubility questions, always check the units requested. If the question asks for $\text{mol dm}^{-3}$, ensure your volume is in $\text{dm}^3$ (liters) and your mass is converted to moles. A common mistake is forgetting to divide by the volume of the solvent—don't let that happen to you!