Answer:
Step-by-step explanation:
Molar concentration of the final solution is the number of moles of solute (n) divided by the final volume in liters (V). This is the mathematical formula for molarity:
M = n solute / V solution in liters = n silver nitrate / V solution in liters
1) Calculate the number of moles of silver nitrate in the initial solution, dissolving 54.0 g of silver nitrate to make 350.0 ml of solution:
- Solute = silver nitrate = AgNO₃
- Molar mass of AgNO₃ = 169.87 g/mol
- n = mass in grams / molar mass = 54.0 g / 169.87 g/mol = 0.318 mol
2) Calculate the molarity of the initial solution:
- M = n / V = 0.318 mol / 0.3500 liter = 0.9086 M
3) Calculate the number of moles of silver nitrate in a 10.0 ml portion of that solution:
n = 0.9086 M × 10.0 ml / (1,000 ml / liter) = 0.009086 moles silver nitrate
4) Molar concentration (molarity) of the final 250.0 ml solution:
- M = n / V = 0.009086 mol / 0.2500 liter = 0.0363 M
5) Concentration of ions present in the final solution:
You must assume 100% ionization:
Then, since each mole of AgNO₃ yields two moles of ions, the total concentration of ions present is the double: 2 × 0.0363 M = 0.0726 M