You have 150.0 mL of a 0.415 M solution of Ce(NO₃)₄.

a) What mass (in grams) of Ce(NO₃)₄ would be required to make the solution?
B) What is the concentration of the nitrate ions in the solution ?

C) If the original solution was diluted to 350.0 mL, what would be the new concentration of the Ce(NO₃)₄ in the solution?

Answers

Answer 1

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 of Ce(NO₃)₄ =?

Mole = Molarity × Volume

Mole of Ce(NO₃)₄ = 0.415 × 0.15

Mole of Ce(NO₃)₄ = 0.06225 mole

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 of Ce(NO₃)₄ =?

Mass = mole × molar mass

Mass of Ce(NO₃)₄ = 0.06225 × 388.12

Mass of Ce(NO₃)₄ = 24.16 g

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) =?

M₁V₁ = M₂V₂

0.415 × 150 = M₂ × 350

62.25 = M₂ × 350

Divide both side by 350

M₂ = 62.25 / 350

M₂ = 0.178 M

Therefore, the new concentration of the Ce(NO₃)₄ solution is 0.178 M

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