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Question: What volume of ​\[0.1M,KMn{O_4}\] is needed to oxidize \(5mg\) of ferrous oxalate in acidic medium? ...

What volume of ​0.1M,KMnO40.1M,KMn{O_4} is needed to oxidize 5mg5mg of ferrous oxalate in acidic medium? (Molecular weight of ferrous oxalate is144g/mol144g/mol)
A.0.20ml0.20ml
B.0.1ml0.1ml
C.0.4ml0.4ml
D.2.08ml2.08ml

Explanation

Solution

One mole needs 55 moles of electrons in an acidic medium to reduce to. To find the volume of 0.1 M0.1{\text{ }}Msolution, which contains the required 0.41670.4167 moles.

Complete step by step answer:
Milliequivalents of KMnO4=V×0.1×5KMn{O_4} = V \times 0.1 \times 5
Milliequivalents of Fe(C2O4)2=Weight(g)Ew×1000Fe{({C_2}{O_4})_2} = \dfrac{{Weight(g)}}{{Ew}} \times 1000
Fe+2Fe+3+eF{e^{ + 2}} \to F{e^{ + 3}} + {e^ - }
C2O422CO2+2e{C_2}O_4^{2 - } \to 2C{O_2} + 2{e^ - }
(n = 3)
Equivalent weight of Fe(C2O4)=Mw3=1443Fe({C_2}{O_4}) = \dfrac{{Mw}}{3} = \dfrac{{144}}{3}
Milliequivalents of Fe(C2O4)=5×1031443×103=5×3144Fe({C_2}{O_4}) = \dfrac{{5 \times {{10}^{ - 3}}}}{{\dfrac{{144}}{3}}} \times {10^3} = \dfrac{{5 \times 3}}{{144}}
Milliequivalents of KMnO4=KMn{O_4} = Milliequivalents of Fe(C2O4)Fe({C_2}{O_4})
V×0.1×5=5×3144\therefore V \times 0.1 \times 5 = \dfrac{{5 \times 3}}{{144}}
V=0.20ml\therefore V = 0.20ml
An equivalent (symbol: officially equivalent; unofficially but often Eq) is the amount of a substance that is (or is equivalent to) an arbitrary quantity of another substance in a given chemical reaction. It is an archaic unit of measurement that was used in chemistry and biological sciences (see equivalent weighting in history). An equivalent mass is called its equivalent weight. In a more formal definition, the equivalent is the amount of substance required to do any of the following: reacting with or supplying one mole of hydrogen ions (H+{H^ + } ) in an acid-base reaction or supplying one mole of electron redox reaction.
According to this definition, the number of moles of an ion in a solution is equal to one, multiplied by the validity of that ion. If 11 mole of NaClNaCl and 11 mole of CaCl2CaC{l_2} dissolve in the solution, that solution contains 11 equiv Na, 22 equiv Ca, and 33 equ ClCl. (Calcium has a validity of 22 , so the ion is equivalent to 11 mole 22.)

Note:
An earlier definition, used exclusively for chemical elements, holds that an equivalent is the amount of a substance that is 1 gram (0.035 oz) of hydrogen, 16 grams (0.56 oz), or 35.5 grams (1.25 oz) Will react with. Chlorine - or that will displace any of the three.