A `20.0mL` sample of `CuSO_(4)` solution was evaporated to dryness, leaving `0.967g` of residue. What was the molarity of the original solution ? `(Cu=63.5)`

Options

A. `48.4M`
B. `0.0207M`
C. `0.0484M`
D. `0.303M`

Explanation

To find the molarity of the original `CuSO₄` solution, we will follow these steps:

### Step 1: Calculate the Molar Mass of `CuSO₄`
The molar mass of `CuSO₄` can be calculated by adding the molar masses of its constituent elements:
- Copper (Cu): 63.5 g/mol
- Sulfur (S): 32.0 g/mol
- Oxygen (O): 16.0 g/mol (and there are 4 oxygen atoms)

Calculating the total:
\[ \text{Molar mass of } CuSO₄ = 63.5 + 32.0 + (4 \times 16.0) = 63.5 + 32.0 + 64.0 = 159.5 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of `CuSO₄`
Using the mass of the residue obtained after evaporation, we can calculate the number of moles of `CuSO₄`:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.967 \text{ g}}{159.5 \text{ g/mol}} \]
Calculating this gives:
\[ \text{Number of moles} = 0.00606 \text{ moles} \]

### Step 3: Convert the Volume of the Solution to Liters
The volume of the solution is given as `20.0 mL`. We need to convert this to liters:
\[ \text{Volume in liters} = \frac{20.0 \text{ mL}}{1000} = 0.020 \text{ L} \]

### Step 4: Calculate the Molarity of the Solution
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity (M)} = \frac{\text{Number of moles}}{\text{Volume in liters}} = \frac{0.00606 \text{ moles}}{0.020 \text{ L}} \]
Calculating this gives:
\[ \text{Molarity (M)} = 0.303 \text{ M} \]

### Final Answer
The molarity of the original `CuSO₄` solution is **0.303 M**.

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