What mass of KBr (in grams) is required to prepare 250.0 mL of a 1.50 M KBr solution?

• A. 28.1 g
• B. 16.5 g
• C. 37.5 g
• D. 44.8 g

Answer: (D)

Verified step by step guidance

1. Identify the given information: volume of solution (V) = 250.0 mL, molarity (M) = 1.50 M. Convert the volume from milliliters to liters because molarity is expressed in moles per liter. Use the conversion: \(V(\text{L}) = \frac{V(\text{mL})}{1000}\).
2. Use the definition of molarity to find the number of moles of KBr needed: \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\). Rearrange to find moles: \(\text{moles} = M \times V\).
3. Calculate the molar mass of KBr by adding the atomic masses of potassium (K) and bromine (Br). Use the periodic table values: approximately 39.1 g/mol for K and 79.9 g/mol for Br. Then, \(\text{molar mass of KBr} = 39.1 + 79.9\) g/mol.
4. Convert moles of KBr to grams using the molar mass: \(\text{mass} = \text{moles} \times \text{molar mass}\).
5. Substitute the values from previous steps into the equation to find the mass of KBr required to prepare the solution.