Chapter 22.7, Problem 3.3ACP

Interpretation Introduction

Interpretation:

Concentration cerium salt in the in the given compound has to be calculated..

 Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

     $\text{Molarity}\text{=}\frac{\text{Mass}\text{per}\text{litre}}{\text{Molecular }\text{mass}}$

• A solution containing one gram mole or $\text{0}\text{.1}$ gram of solute per litre of solution is called molar solution.
• Amount of substance (mol) can be determined by using the equation,                      

    $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• $\text{Concentration}\text{of}\text{substance}\text{=}\frac{\text{Amount}\text{of substance}}{\text{volume}\text{of}\text{the}\text{substance}}$
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

Answer to Problem 3.3ACP

Concentration cerium salt in the in the given compound is $\text{0}\text{.103}\text{M}$

Explanation of Solution

Probable equation for the given reaction can be written as follows,

$F{e}_{\left(aq\right)}^{2+}+{\left[Ce{\left(N{O}_3\right)}_6\right]}_{\left(aq\right)}^{2-}\to F{e}_{\left(aq\right)}^{3+}+C{e}_{\left(aq\right)}^{3+}+6N{O}_{3\left(aq\right)}^{-}$

Given mass of $Feis0.181g$ which is dissolved in acid, reduced to $F{e}^{2+}$ and then titrated with $31.33mLof{\left(N{H}_4\right)}_{2}Ce{\left(N{O}_3\right)}_6$

Amount of ${\text{Fe}}^{\text{2+}}$ (in mol) can be calculated as follows,

                      $0.181\text{g}\text{Fe}\text{×}\frac{\text{1}\text{mol}\text{of}\text{Fe}}{55.85\text{g}\text{Fe}}\text{=}\text{0}\text{.00324}\text{mol}\text{Fe}$

${\text{Fe}}^{\text{2+}}$ and cerium salt in the reaction are in $1:1$ ratio. Therefore, from the amount of ${\text{Fe}}^{\text{2+}}$, concentration cerium salt can be calculated as follows,

$\begin{array}{c}AmountofCesalt=0.00324molFe\times \frac{1mol{\left[Ce{\left(N{O}_3\right)}_6\right]}^{2-}}{1molFe}\\ =0.00324molCesalt\end{array}$

$\text{Concentration}\text{of}\text{Ce}\text{salt}\text{=}\frac{\text{0}\text{.00324}\text{mol}\text{Ce}\text{salt}}{\text{0}\text{.03133}\text{L}}\text{=}\text{0}\text{.103}\text{M}$

Conclusion

Concentration cerium salt in the in the given compound was determined.