Question

1. find the mass of iron (ii) sulphate needed to prepare 250ml of 0.600mol/l solution. a-5
2. copper is an element that is required in very small concentrations in our body. what is the concentration, in parts per million, if 1.0l of drinking water contains 3.0×10⁻⁵g of copper? a-3

Explanation:

Problem 4: Find the mass of iron (II) sulphate needed to prepare 250 mL of 0.600 mol/L solution.

Step 1: Convert volume to liters

The volume of the solution is 250 mL. Since 1 L = 1000 mL, we convert 250 mL to liters:
\( V = \frac{250\ \text{mL}}{1000\ \text{mL/L}} = 0.250\ \text{L} \)

Step 2: Calculate moles of iron (II) sulphate

Using the formula \( n = C \times V \), where \( C = 0.600\ \text{mol/L} \) and \( V = 0.250\ \text{L} \):
\( n = 0.600\ \text{mol/L} \times 0.250\ \text{L} = 0.150\ \text{mol} \)

Step 3: Determine molar mass of \( \text{FeSO}_4 \)

Iron (Fe) has a molar mass of \( 55.85\ \text{g/mol} \), sulfur (S) is \( 32.07\ \text{g/mol} \), and oxygen (O) is \( 16.00\ \text{g/mol} \). For \( \text{FeSO}_4 \):
\( M = 55.85 + 32.07 + (4 \times 16.00) = 55.85 + 32.07 + 64.00 = 151.92\ \text{g/mol} \) (Note: The original work used 162.2 g/mol, likely a rounded value for \( \text{FeSO}_4 \cdot \text{H}_2\text{O} \) or similar, but we follow the calculation logic.)

Step 4: Calculate mass using \( m = n \times M \)

\( m = 0.150\ \text{mol} \times 151.92\ \text{g/mol} \approx 22.79\ \text{g} \) (If using \( M = 162.2\ \text{g/mol} \) as in the original: \( m = 0.150\ \text{mol} \times 162.2\ \text{g/mol} = 24.33\ \text{g} \))

Problem 5: Concentration of copper in parts per million (ppm)

Step 1: Mass of solution (water)

Assume the density of water is \( 1.0\ \text{g/mL} \). For 1.0 L (1000 mL) of water:
\( m_{\text{solution}} = 1000\ \text{mL} \times 1.0\ \text{g/mL} = 1000\ \text{g} \)

Step 2: Calculate ppm using \( \text{ppm} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 10^6 \)

Given \( m_{\text{solute}} = 3.0 \times 10^{-5}\ \text{g} \) and \( m_{\text{solution}} = 1000\ \text{g} \):
\( \text{ppm} = \frac{3.0 \times 10^{-5}\ \text{g}}{1000\ \text{g}} \times 10^6 = \frac{3.0 \times 10^{-5} \times 10^6}{1000} = \frac{30}{1000} = 0.03\ \text{ppm} \)

Answer:

s:

1. The mass of iron (II) sulphate is \( \boldsymbol{24.33\ \text{g}} \) (using \( M = 162.2\ \text{g/mol} \)) or \( \boldsymbol{22.8\ \text{g}} \) (using anhydrous \( \text{FeSO}_4 \)).
2. The concentration of copper is \( \boldsymbol{0.03\ \text{ppm}} \).