Question

How many grams of $ \mathrm{Al_2(SO_4)_3} $ are required to produce 10 L of a 0.5 M solution? (Molar mass = 342 g/mol)

• A. 1710 g
• B. 342 g
• C. 68.4 g
• D. 85.5 g

Hint:

Use the formula: \[ \text{Mass} = Molarity \times Volume \times Molar\, mass \] Ensure all units are compatible: Volume in Liters and molar mass in g/mol.

Answer 1

The Correct Option is A

To determine how many grams of aluminum sulfate, \( \mathrm{Al_2(SO_4)_3} \), are required to make a 10 L solution at 0.5 M concentration, follow these steps:

Step 1: Understand Molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula is:

\[\text{Molarity (M)}=\frac{\text{moles of solute}}{\text{volume of solution in liters}}\]

Given a 0.5 M solution, this means we have 0.5 moles of \( \mathrm{Al_2(SO_4)_3} \) per liter. For 10 L, we calculate:

Step 2: Calculate Moles of \( \mathrm{Al_2(SO_4)_3} \)
\(\text{Moles} = 0.5 \text{ M} \times 10 \text{ L} = 5 \text{ moles}\)

Step 3: Convert Moles to Grams
Using the molar mass of \( \mathrm{Al_2(SO_4)_3} \), which is 342 g/mol, convert the moles to grams:

\(\text{Mass (g)} = \text{Moles} \times \text{Molar Mass (g/mol)}\)

Substitute the known values:

\(\text{Mass} = 5 \text{ moles} \times 342 \text{ g/mol} = 1710 \text{ g}\)

Conclusion:
1710 grams of \( \mathrm{Al_2(SO_4)_3} \) are required to make 10 L of a 0.5 M solution.

Answer 2

Step 1: Use the molarity formula 
\[ M = \frac{n}{V} \] Where: 
\( M \) is the molarity in mol/L 
\( n \) is the number of moles 
\( V \) is the volume in liters
Step 2: Rearrange to find number of moles \[ n = M \times V \] Substitute values: \[ n = 0.5 \, \text{mol/L} \times 10 \, \text{L} = 5 \, \text{mol} \] Step 3: Use molar mass to convert moles to grams \[ \text{Mass} = n \times \text{Molar mass} = 5 \, \text{mol} \times 342 \, \text{g/mol} = 1710 \, \text{g} \] Final Answer: $\boxed{1710 g}$