Calculate the solubility of solid X in g/100g H2O for volumes 3, 4, 6, and 8 cm3 of water, given 4.5g of X.

Here are the calculations and the completed table for the solubility of X, along with an explanation for plotting the graph and addressing the other parts of the question.

To complete Table I, we need to calculate the solubility of solid X for each volume of distilled water used. The mass of solid X is given as $4.5 \text{ g}$. We assume the density of water is $1 \text{ g/cm}^3$, so the volume of water in $\text{cm}^3$ is equal to its mass in grams.

The formula for solubility is:

$$\text{Solubility (g/100g H}_2\text{O)} = \frac{\text{Mass of X (g)}}{\text{Mass of H}_2\text{O (g)}} \times 100$$

Step 1: Calculate the solubility for each volume of water.

• For $3 \text{ cm}^3$ of water: Mass of $\text{H}_2\text{O} = 3 \text{ g}$

  $$\text{Solubility} = \frac{4.5 \text{ g}}{3 \text{ g}} \times 100 = 1.5 \times 100 = 150 \text{ g/100g H}_2\text{O}$$

• For $4 \text{ cm}^3$ of water: Mass of $\text{H}_2\text{O} = 4 \text{ g}$

  $$\text{Solubility} = \frac{4.5 \text{ g}}{4 \text{ g}} \times 100 = 1.125 \times 100 = 112.5 \text{ g/100g H}_2\text{O}$$

• For $6 \text{ cm}^3$ of water: Mass of $\text{H}_2\text{O} = 6 \text{ g}$

  $$\text{Solubility} = \frac{4.5 \text{ g}}{6 \text{ g}} \times 100 = 0.75 \times 100 = 75 \text{ g/100g H}_2\text{O}$$

• For $8 \text{ cm}^3$ of water: Mass of $\text{H}_2\text{O} = 8 \text{ g}$

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