When 0.083 moles of lead(II) nitrate are dissolved in enough water to make 617 mililiters
of solution, what is the molar concentration
of nitrate ions?
Answer in units of
Answer in units of m. Whoopsies~
To find the molar concentration of nitrate ions in the solution, we need to know the number of moles of nitrate ions and the volume of the solution in liters.
First, we need to determine the number of moles of nitrate ions present in 0.083 moles of lead(II) nitrate.
Lead(II) nitrate has the chemical formula Pb(NO3)2, which means there are two moles of nitrate ions (NO3-) for every mole of lead(II) nitrate (Pb(NO3)2).
Since we have 0.083 moles of lead(II) nitrate, we can calculate the number of moles of nitrate ions by multiplying that by the ratio of 2 moles of NO3- per 1 mole of Pb(NO3)2:
0.083 moles Pb(NO3)2 * (2 moles NO3- / 1 mole Pb(NO3)2) = 0.166 moles NO3-
Now, we need to convert the volume of the solution from milliliters to liters. The given volume is 617 milliliters, which is equivalent to 617/1000 = 0.617 liters.
Finally, we can calculate the molar concentration of nitrate ions by dividing the number of moles of nitrate ions by the volume of the solution in liters:
Molar concentration = 0.166 moles NO3- / 0.617 liters = 0.269 M
Therefore, the molar concentration of nitrate ions in the solution is 0.269 M (moles per liter).