Question

5.5 g of a mixture of \[{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\] and \[{\rm{F}}{{\rm{e}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{9}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\] requires 5.4 ml of 0.1 N \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\] solution for complete oxidation. Calculate the number of moles of \[{\rm{F}}{{\rm{e}}_{\rm{2}}}\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right){\rm{39}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\] in the mixture.
A. 0.0095
B. 0.15
C. 0.0952
D. 1.52

Answer 1

Hint: The term milliequivalent is the unit used to measure the concentration of electrolytes. It measures the chemical reactivity of an electrolyte.

Formula used:
The formula to calculate mole number is,
Number of moles=\[\dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\,{\rm{mass}}}}\]

Complete Step by Step Solution:
Here, we have to first calculate the mass of \[{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\].s

The reactions are,
\[5{e^ - } + {\rm{M}}{{\rm{n}}^{{\rm{ + 7}}}} \to {\rm{M}}{{\rm{n}}^{{\rm{ + 2}}}}\]
\[{\rm{F}}{{\rm{e}}^{{\rm{2 + }}}} \to {\rm{F}}{{\rm{e}}^{{\rm{3 + }}}} + {e^ - }\]
So, only \[{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]undergoes reaction with \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]. Therefore, miliequivalents of \[{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]is equal to miliequivalent of \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\].
\[{{\rm{M}}_{{\rm{eq}}}}\,{\rm{of}}\,{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}} = \,{{\rm{M}}_{{\rm{eq}}}}\,{\rm{of}}\,{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\] ---- (1)

The formula of milliequivalent is,
mEq=\[\dfrac{{{\rm{Weight}}}}{{{\rm{Eq}}{\rm{.}}\,{\rm{weight}}}}{\rm{ \times 1000}}\]
So, equation (1) becomes,
\[\dfrac{{{\rm{Weight}}}}{{{\rm{Eq}}{\rm{.}}\,{\rm{weight}}}}{\rm{ \times 1000}} = 0.1 \times 5.4 \times 1\] ----- (2)
Equivalent weight formula is \[\dfrac{{{\rm{Molecular}}\,{\rm{formula}}}}{{{\rm{Valency}}\,{\rm{factor}}}}\]

Weight of \[{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]is 178 g and the valency factor for \[{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]is 1. So, Equation (2) becomes,
\[\dfrac{{{\rm{Weight}}}}{{\dfrac{{178}}{1}}}{\rm{ \times 1000}} = 0.1 \times 5.4 \times 1\]
\[{{\rm{W}}_{{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = {\rm{0}}{\rm{.150}}\,{\rm{g}}\]

The total mass is given 5.5 g and the mass of \[{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{.7}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]is 0.150 g. Now, we have to calculate the mass of \[{\rm{F}}{{\rm{e}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{9}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\].
Mass of \[{\rm{F}}{{\rm{e}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{9}}{{\rm{H}}_{\rm{2}}}{\rm{O}} = 5.5 - 0.150 = 5.35\,{\rm{g}}\]

Now, we have to find out the moles of \[{\rm{F}}{{\rm{e}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{9}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]. The molar mass of \[{\rm{F}}{{\rm{e}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{9}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]is 562 g.
Number of moles=\[\dfrac{{5.35}}{{562}} = 0.00951\,{\rm{mol}}\]

Therefore, the moles of \[{\rm{F}}{{\rm{e}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{9}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]in the mixture is 0.00951.
Hence, option (c) is correct.

Note: Equivalent weight defines the mass of a substance that undergoes combination with a fixed mass of another substance. Originally equivalent weights were calculated experimentally, but nowadays they are calculated from molar masses. A compound's equivalent weight is calculated by the division of molecular mass by the quantity of charge (negative or positive) resulting from the dissolution of the compound.