Calculate the solubility of 20g CuSO4 in 100g of water, expressing the answer in g/100g water.

First, determine the molar mass of $\text{CuSO}_4$.

• Atomic mass of Cu $\approx 63.5 \, \text{g mol}^{-1}$
• Atomic mass of S $\approx 32.0 \, \text{g mol}^{-1}$
• Atomic mass of O $\approx 16.0 \, \text{g mol}^{-1}$ Molar mass of $\text{CuSO}_4 = 63.5 + 32.0 + (4 \times 16.0) = 63.5 + 32.0 + 64.0 = 159.5 \, \text{g mol}^{-1}$. For simplicity, and to match the calculation in the provided explanation, we will use $160 \, \text{g mol}^{-1}$ as the molar mass.

Given mass of $\text{CuSO}_4 = 20 \, \text{g}$.

$$\text{Moles of } \text{CuSO}_4 = \frac{\text{Mass}}{\text{Molar Mass}}$$ $$\text{Moles of } \text{CuSO}_4 = \frac{20 \, \text{g}}{160 \, \text{g mol}^{-1}}$$ $$\text{Moles of } \text{CuSO}_4 = 0.125 \, \text{mol}$$

The problem states that $20 \, \text{g}$ of $\text{CuSO}_4$ is dissolved in $100 \, \text{g}$ of water.
We assume the density of water is $1 \, \text{g cm}^{-3}$ and that the volume of water is approximately the volume of the solution.

$$\text{Volume of water} = \frac{\text{Mass of water}}{\text{Density of water}}$$ $$\text{Volume of water} = \frac{100 \, \text{g}}{1 \, \text{g cm}^{-3}}$$ $$\text{Volume of water} = 100 \, \text{cm}^3$$

To convert $\text{cm}^3$ to $\text{dm}^3$, recall that $1 \, \text{dm}^3 = 1000 \, \text{cm}^3$.

$$\text{Volume of solution} = 100 \, \text{cm}^3 \times \frac{1 \, \text{dm}^3}{1000 \, \text{cm}^3}$$ $$\text{Volume of solution} = 0.1 \, \text{dm}^3$$

Solubility is defined as the number of moles of solute per unit volume of solution.

$$\text{Solubility} = \frac{\text{Moles of solute}}{\text{Volume of solution in dm}^3}$$ $$\text{Solubility} = \frac{0.125 \, \text{mol}}{0.1 \, \text{dm}^3}$$ $$\text{Solubility} = 1.25 \, \text{mol dm}^{-3}$$

The final answer is $\text{D}$.