Final answer:
The concentration of Cl− ions after dissolving 3.50 g NaCl in 50.00 mL of 0.500 M CaCl₂ solution and considering the molarity of ions is 2.198 M, which is not an option in the question. There may be a typo as the answer should be close to option C (1.750 M) but is calculated to be higher than any options given.
Step-by-step explanation:
To determine the concentration of Cl− ions after dissolving 3.50 g NaCl in 50.00 mL of 0.500 M CaCl₂ solution, we need to calculate the moles of Cl− from both compounds and then find the total concentration.
First, let's find the moles of Cl− from NaCl:
- Molar mass of NaCl = 58.44 g/mol
- Moles of NaCl = 3.50 g / 58.44 g/mol = 0.0599 mol
- Since NaCl dissociates into Na⁺ and Cl−, the moles of Cl− from NaCl will also be 0.0599 mol.
Now, let's look at the CaCl₂ solution:
- Concentration of CaCl₂ = 0.500 M
- Each CaCl₂ provides two Cl− ions upon dissociation.
- Moles of CaCl₂ in 50.00 mL = 0.500 M × 0.050 L = 0.025 mol
- Moles of Cl− from CaCl₂ = 0.025 mol CaCl₂ × 2 = 0.050 mol
Summing up the moles of Cl− ions:
- Total moles of Cl− = 0.0599 mol from NaCl + 0.050 mol from CaCl₂ = 0.1099 mol
Calculating the resultant concentration of Cl−:
- Volume of the final solution = 50.00 mL of CaCl₂ + volume of NaCl (assuming negligible volume change) = ≈ 50.00 mL
- Total volume in liters = 50.00 mL / 1000 mL/L = 0.05 L
- Concentration of Cl− = Total moles of Cl− / Total volume in liters = 0.1099 moles / 0.05 L = 2.198 M