A. The mass (in grams) of Ce(NO₃)₄ that will be required to make the solution is 24.16 g
B. The concentration of the nitrate ions in the solution is 1.66 M
C. The new concentration of the Ce(NO₃)₄ solution is 0.178 M
We'll begin by calculating the number of mole of Ce(NO₃)₄ in the solution.
Molarity = 0.415 M
Volume = 150 mL = 150 /1000 = 0.15 L
Mole = Molarity × Volume
Mole of Ce(NO₃)₄ = 0.415 × 0.15
A. Determination of the mass of Ce(NO₃)₄ required to make the solution.
Mole of Ce(NO₃)₄ = 0.06225 mole
Molar of Ce(NO₃)₄ = 140.12 + 4[14 + (16×3)]
= 140.12 + 4[14 + 48]
= 140.12 + 4[62]
= 140.12 + 248
= 388.12 g/mol
Mass = mole × molar mass
Mass of Ce(NO₃)₄ = 0.06225 × 388.12
Therefore, 24.16 g of Ce(NO₃)₄ is required to make the solution.
B. Determination of the concentration of the nitrate ion, NO₃¯ in the solution.
In solution, Ce(NO₃)₄ will dissociate as follow:
Ce(NO₃)₄ (aq) —> Ce⁴⁺(aq) + 4NO₃¯(aq)
From the balanced equation above,
1 mole of Ce(NO₃)₄ contains 4 moles of NO₃¯
Therefore,
0.415 M solution of Ce(NO₃)₄ will contain = 0.415 × 4 = 1.66 M NO₃¯
Thus, 1.66 M of the nitrate ion, NO₃¯, is present in the solution.
C. Determination of the new concentration of the Ce(NO₃)₄ solution.
Volume of stock solution (V₁) = 150 mL
Molarity of stock solution (M₁) = 0.415 M
Volume of diluted solution (V₂) = 350 mL
Molarity of diluted solution (M₂) =?
0.415 × 150 = M₂ × 350
62.25 = M₂ × 350
Divide both side by 350
M₂ = 62.25 / 350
Therefore, the new concentration of the Ce(NO₃)₄ solution is 0.178 M
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