What mass of KBr is required to prepare 35.0 mL of a 1.80 M KBr solution?

• A. 7.48 g
• B. 2.10 g
• C. 1.80 g
• D. 10.5 g

Answer: (A)

Verified step by step guidance

1. Identify the given information: volume of solution (V) = 35.0 mL and molarity (M) = 1.80 M. Remember to convert the volume from milliliters to liters because molarity is expressed in moles per liter.
2. Convert the volume from milliliters to liters using the conversion factor: \(\text{Volume (L)} = \frac{35.0\ \text{mL}}{1000}\).
3. Use the molarity definition to find the number of moles of KBr needed: \(\text{moles of KBr} = M \times V = 1.80\ \text{mol/L} \times \text{Volume (L)}\).
4. Calculate the molar mass of KBr by adding the atomic masses of potassium (K) and bromine (Br). Use the periodic table values: approximately 39.10 g/mol for K and 79.90 g/mol for Br.
5. Find the mass of KBr required by multiplying the moles of KBr by its molar mass: \(\text{mass} = \text{moles} \times \text{molar mass}\).