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 ascertain the mass in grams of aluminum sulfate, \( \mathrm{Al_2(SO_4)_3} \), needed for a 10 L solution at 0.5 M concentration, execute the subsequent steps:

Step 1: Grasp Molarity Definition
Molarity (M) quantifies solute moles per solution liter. The governing equation is:

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

A 0.5 M solution signifies 0.5 moles of \( \mathrm{Al_2(SO_4)_3} \) per liter. For 10 L, the moles are computed as:

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

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

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

Input the established values:

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

Outcome:
1710 grams of \( \mathrm{Al_2(SO_4)_3} \) are requisite for the preparation of 10 L of a 0.5 M solution.