Determine the molarity of nitrate ions in a solution prepared by mixing 25.0 mL of 0.50 M Fe(NO3)3 and 35.0 mL of 1.00 M Mg(NO3)2 Please explain fully do not know where to start?
You start with the definition of molarity. M = moles/L.
moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.
moles Mg(NO3)2 = M x L = ??
moles NO3^- will be 2x that.
Then M NO3^- = total moles NO3^-/total volume in L.
(0.50)(25.0) = 12.5
(1.00)(35.0) = 35
47.5/60 = 0.79M is my answer but my book says
1.79M is the correct answer
You didn't follow any of my instructions. I have typed my bolded instructions below each of your lines.
0.50)(25.0) = 12.5
moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.
M x L = 0.50M x 0.025 L(not 25 mL) = 0.0125 moles Fe(NO3)3
Then nitrate is 3x that or 0.0125*3 = 0.03750 moles nitrate ion
(1.00)(35.0) = 35
moles = M x L = 1.00M x 0.0350 L (not mL) = 0.0350 moles Mg(NO3)2.
Then nitrate is twice that or
2*0.0350 = 0.0700 M in nitrate.
47.5/60 = 0.79M is my answer but my book says
M nitrate = total moles/total volume in L. (0.0375+0.0700)/0.06L = 1.79 M in nitrate.
1.79M is the correct answer
So it is
so I times 0.0125 by 3 b/c of the 3 at the end of the bracket?
and I times 0.0350 by 2 b/c of the 2 at the end of the bracket?
but why because when i'm doing my molar mass i do include that last # in the molar mass so why do we times it by that #?
To determine the molarity of nitrate ions in the solution, we need to calculate the moles of nitrate ions in both compounds and then determine the total moles of nitrate ions in the final solution.
First, let's start by calculating the number of moles of nitrate ions in Fe(NO3)3 and Mg(NO3)2.
1) Fe(NO3)3:
Molarity of Fe(NO3)3 = 0.50 M
Volume of Fe(NO3)3 = 25.0 mL = 0.025 L
Moles of Fe(NO3)3 = molarity × volume
= 0.50 M × 0.025 L
= 0.0125 moles
Fe(NO3)3 dissociates into one Fe3+ ion and three NO3- ions.
Therefore, the moles of nitrate ions in Fe(NO3)3 = 3 × 0.0125 moles
= 0.0375 moles
2) Mg(NO3)2:
Molarity of Mg(NO3)2 = 1.00 M
Volume of Mg(NO3)2 = 35.0 mL = 0.035 L
Moles of Mg(NO3)2 = molarity × volume
= 1.00 M × 0.035 L
= 0.035 moles
Mg(NO3)2 dissociates into one Mg2+ ion and two NO3- ions.
Therefore, the moles of nitrate ions in Mg(NO3)2 = 2 × 0.035 moles
= 0.07 moles
Now that we have calculated the moles of nitrate ions in both compounds, we can determine the total moles of nitrate ions in the final solution by adding the moles obtained from Fe(NO3)3 and Mg(NO3)2 together.
Total moles of nitrate ions = Moles from Fe(NO3)3 + Moles from Mg(NO3)2
= 0.0375 moles + 0.07 moles
= 0.1075 moles
Finally, to calculate the molarity of nitrate ions in the final solution, we need to divide the total moles by the total volume of the solution.
Total volume of the solution = Volume of Fe(NO3)3 + Volume of Mg(NO3)2
= 25.0 mL + 35.0 mL
= 60.0 mL = 0.060 L
Molarity of nitrate ions = Total moles of nitrate ions / Total volume of the solution
= 0.1075 moles / 0.060 L
= 1.79 M
Therefore, the molarity of nitrate ions in the solution prepared by mixing 25.0 mL of 0.50 M Fe(NO3)3 and 35.0 mL of 1.00 M Mg(NO3)2 is approximately 1.79 M.