Final answer:
To find the required grams of CaCl₂ for a 4.55 M solution, multiply the volume in liters by molarity to find moles, then moles by molar mass to get grams. The answer is approximately 270 grams.
Step-by-step explanation:
The question asks how many grams of CaCl₂ are needed to make 535 mL of 4.55 M CaCl₂ solution. To begin, we need to use the molarity equation: Molarity (M) = moles of solute / liters of solution. We can solve for moles and then for grams.
- First, convert 535 mL to liters by dividing by 1000. That gives us 0.535 liters.
- Next, multiply the volume in liters (0.535 L) by the molarity (4.55 M) to get the moles of CaCl₂ needed: 0.535 L × 4.55 M = 2.43425 moles.
- Last, multiply the moles of CaCl₂ by its molar mass (110.98 g/mol) to find the mass in grams: 2.43425 moles × 110.98 g/mol = 270.04 grams of CaCl₂.
Hence, the answer is 270 grams of CaCl₂ (rounded to the nearest whole number), which corresponds to option e.