
What weight of ferrous ammonium sulphate is needed to prepare $ 100{\text{ }}ml $ of $ 0.1 $ normal solution $ (molecular{\text{ weight 392)}} $ .
A. $ 39.2{\text{ gm}} $
B. $ 3.92{\text{ gm}} $
C. $ 1.96{\text{ gm}} $
D. $ 19.6{\text{ gm}} $
Answer
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Hint: Ferrous ammonium sulphate is an inorganic compound also known as Mohr’s salt. Ferrous ammonium sulphate is a double salt of ferrous sulphate and ammonium sulphate, moreover this inorganic compound contains two different cations i.e. ferrous sulphate and ammonium sulphate. Mohr’s salt is a laboratory reagent because of its property of crystallization and the resistance of the crystals towards oxidation by the air.
Complete answer:
Chemical formula of ferrous ammonium sulphate is $ FeS{O_4}(N{H_4})S{O_4}.6{H_2}O $ .
$ F{e^{2 + }} $ is oxidised to $ F{e^{3 + }} $ , but nitrogen and sulphur is in the highest oxidation state.
$ FeS{O_4} \to F{e^{3 + }} + {e^ - } $
Equivalent weight is defined as the ratio of the molecular weight of the molecular weight of the solute to the valency factor of the solute. The equivalent mass of an acid is equal to the ratio of the molecular weight of the acid by its basicity, whereas the equivalent mass of a base is equal to the molecular mass of the base by its acidity.
Equivalent mass $ = \dfrac{{molecular{\text{ mass}}}}{n} $
Normality, which is also known as equivalent concentration of a solution, is defined as the ratio of molar concentration to the equivalence factor. In standard form, normality is the number of grams or mole equivalents of solute present in one litre of a solution.
Therefore Normality $ = \dfrac{{\operatorname{Re} quired{\text{ mass}}}}{{Equivalent{\text{ mass}}}} \times \dfrac{{1000}}{{Volume{\text{ of solution (in mL)}}}} $
$ 0.1 = \dfrac{x}{{392}} \times \dfrac{{1000}}{{100}} $
Making $ x $ the subject of the equation
$ x = 392 \times 0.1 \times \dfrac{1}{{10}} $
$ x = 3.92g $
$ 3.92 $ grams of ferrous ammonium sulphate is needed to prepare $ 100 ml $ of $ 0.1 $ normal solution.
Therefore the correct answer is option B.
Note:
Molar concentration also known as molarity is defined as the number of moles of solute in one litre of solution. Molar concentration is a quantitative way of describing the concentration of solute in a solution. The unit of molarity or molar concentration is moles per litre. Molar concentration depends upon the volume of the solution.
Complete answer:
Chemical formula of ferrous ammonium sulphate is $ FeS{O_4}(N{H_4})S{O_4}.6{H_2}O $ .
$ F{e^{2 + }} $ is oxidised to $ F{e^{3 + }} $ , but nitrogen and sulphur is in the highest oxidation state.
$ FeS{O_4} \to F{e^{3 + }} + {e^ - } $
Equivalent weight is defined as the ratio of the molecular weight of the molecular weight of the solute to the valency factor of the solute. The equivalent mass of an acid is equal to the ratio of the molecular weight of the acid by its basicity, whereas the equivalent mass of a base is equal to the molecular mass of the base by its acidity.
Equivalent mass $ = \dfrac{{molecular{\text{ mass}}}}{n} $
Normality, which is also known as equivalent concentration of a solution, is defined as the ratio of molar concentration to the equivalence factor. In standard form, normality is the number of grams or mole equivalents of solute present in one litre of a solution.
Therefore Normality $ = \dfrac{{\operatorname{Re} quired{\text{ mass}}}}{{Equivalent{\text{ mass}}}} \times \dfrac{{1000}}{{Volume{\text{ of solution (in mL)}}}} $
$ 0.1 = \dfrac{x}{{392}} \times \dfrac{{1000}}{{100}} $
Making $ x $ the subject of the equation
$ x = 392 \times 0.1 \times \dfrac{1}{{10}} $
$ x = 3.92g $
$ 3.92 $ grams of ferrous ammonium sulphate is needed to prepare $ 100 ml $ of $ 0.1 $ normal solution.
Therefore the correct answer is option B.
Note:
Molar concentration also known as molarity is defined as the number of moles of solute in one litre of solution. Molar concentration is a quantitative way of describing the concentration of solute in a solution. The unit of molarity or molar concentration is moles per litre. Molar concentration depends upon the volume of the solution.
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