Question

13.00 grams of Fe(NO3)3 are dissolved in enough water to prepare 400.0ml of solution. How many...

13.00 grams of Fe(NO3)3 are dissolved in enough water to prepare 400.0ml of solution. How many ml of this solution would be be required to supply 0.134 moles of nitrate ions?

Homework Answers

Answer #1

first find the Molarity of the given solution

Molar mass of Fe(NO3)3,
MM = 1*MM(Fe) + 3*MM(N) + 9*MM(O)
= 1*55.85 + 3*14.01 + 9*16.0
= 241.88 g/mol
mass(Fe(NO3)3)= 13.00 g
number of mol of Fe(NO3)3,
n = mass of Fe(NO3)3/molar mass of Fe(NO3)3
=(13.0 g)/(241.88 g/mol)
= 5.375*10^-2 mol
volume , V = 400.0 mL
= 0.4 L
Molarity,
M = number of mol / volume in L
= 5.375*10^-2/0.4
= 0.1344 M

This is molarity of initial solution

Now we need 0.134 moles
use:
number of moles = Molarity * volume
0.134 moles = 0.01344 M * volume
volume = 9.97 L
Answer: 9.97 L

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