How many grams of KCl are required to prepare 50.0 mL of a 2.45 M KCl solution?

• A. 9.13 g
• B. 12.25 g
• C. 6.12 g
• D. 4.58 g

Answer: (A)

Verified step by step guidance

1. Identify the given information: volume of solution (V) = 50.0 mL, molarity (M) = 2.45 M, and the solute is KCl.
2. Convert the volume from milliliters to liters because molarity is expressed in moles per liter: \(V (L) = \frac{50.0\ \text{mL}}{1000}\).
3. Use the molarity definition to find the moles of KCl needed: \(\text{moles of KCl} = M \times V (L) = 2.45 \times V (L)\).
4. Calculate the molar mass of KCl by adding the atomic masses of potassium (K) and chlorine (Cl): \(\text{molar mass of KCl} = 39.10 + 35.45\) g/mol.
5. Find the mass of KCl required by multiplying the moles by the molar mass: \(\text{mass} = \text{moles} \times \text{molar mass}\).