When 0.03 moles of lead(II) nitrate are dissolved in enough water to make 266 milliliters of solution, what is the molar concentration of nitrate ions?
Answer in units of M
I already tried moles per liter and it says it is wrong.
Now that you know what you did wrong on the other two you know what yu did wrong on this one and you should be able to correct it yourself. You aren't reading the problem(s) correctly.
To find the molar concentration of nitrate ions, you need to determine the number of moles of nitrate ions present in the solution and then divide it by the volume of the solution in liters. Let's break it down step by step.
1. Start by calculating the number of moles of nitrate ions (NO3-) present in 0.03 moles of lead(II) nitrate (Pb(NO3)2). Since each lead(II) nitrate molecule dissociates into one Pb2+ ion and two NO3- ions, the number of moles of nitrate ions will be twice the moles of lead(II) nitrate in this case.
Moles of NO3- = 2 × 0.03 moles = 0.06 moles
2. Next, convert the volume of the solution from milliliters to liters. There are 1000 milliliters in a liter, so:
Volume of solution in liters = 266 milliliters / 1000 = 0.266 liters
3. Finally, divide the number of moles of nitrate ions by the volume of the solution in liters to calculate the molar concentration of nitrate ions.
Molar concentration of nitrate ions (NO3-) = Moles of NO3- / Volume of solution in liters
Molar concentration of nitrate ions = 0.06 moles / 0.266 liters ≈ 0.226 M
Therefore, the molar concentration of nitrate ions in the solution is approximately 0.226 M. Make sure to round your final answer to the appropriate number of significant figures.