A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100.0 mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275.0 mL of a final solution. How many grams of NO3- ion are there in the final solution?

• A. 0.024 g
• B. 0.048 g
• C. 0.192 g
• D. 0.012 g

Answer: (B)

Verified step by step guidance

1. First, calculate the molar mass of Co(NO3)2. Use the atomic masses: Co = 58.93 g/mol, N = 14.01 g/mol, and O = 16.00 g/mol. The formula is Co(NO3)2, which means it contains one Co atom, two N atoms, and six O atoms.
2. Determine the number of moles of Co(NO3)2 in the 4.00 g of the compound using the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
3. Calculate the concentration of the stock solution in mol/L (Molarity) using the formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). Remember to convert 100.0 mL to liters.
4. Use the dilution formula \( C_1V_1 = C_2V_2 \) to find the concentration of the final solution after dilution. Here, \( C_1 \) is the concentration of the stock solution, \( V_1 \) is the volume of the stock solution taken (4.00 mL), \( C_2 \) is the concentration of the final solution, and \( V_2 \) is the final volume (275.0 mL).
5. Calculate the mass of NO3- ions in the final solution. Since each Co(NO3)2 molecule contains two NO3- ions, multiply the moles of Co(NO3)2 in the final solution by 2 to get the moles of NO3-. Then, convert moles of NO3- to grams using the molar mass of NO3- (62.01 g/mol).