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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Redox Reactions

NTA Abhyas 2020 · Chemistry · Redox Reactions

Volume of 0.1 M K _ 2 C r _ 2 O _ 7 required to oxidize 35 mL of...

Volume of 0.1 M K2Cr2O7{K}_{2}C{r}_{2}{O}_{7} required to oxidize 35 mL of 0.5 M FeSO4FeS{O}_{4} solution is

Options

  1. A29.2 mL
  2. B175 mL
  3. C185 mL
  4. D145 mL

Correct answer

A. 29.2 mL

Step-by-step solution

Cr2O72−+14H++6e−→2Cr3++7H2OC{r}_{2}{O}_{7}^{2 -} + 14H^{+} + 6e^{-} \to 2Cr^{3 +} + 7{H}_{2}O
a1=6{a}_{1} = 6
a2=1{a}_{2} = 1
a1=n{a}_{1} = n -factor of K2Cr2O7{K}_{2}C{r}_{2}{O}_{7}
a2=n{a}_{2} = n -factor of FeSO4FeS{O}_{4}
Fe2+→Fe3++e−Fe^{2 +} \to Fe^{3 +} + e^{-}
a1M1V1(K2Cr2O7)a1=6=a2M2V2(FeSO4)a2=1\mathop{{a}_{1}{M}_{1}{V}_{1}}_{\mathop{\left({K}_{2}C{r}_{2}{O}_{7}\right)}_{{a}_{1} = 6}} = \mathop{{a}_{2}{M}_{2}{V}_{2}}_{\mathop{\left(FeS{O}_{4}\right)}_{{a}_{2} = 1}}
M1=0.1M{M}_{1} = 0.1\text{M}
M2=0.5{M}_{2} = 0.5 M
V1=?{V}_{1} = ?
V2=35{V}_{2} = 35 mL
So V1=1×0.5×35mL6×0.1{V}_{1} = \dfrac{1 \times 0.5 \times 35\text{mL}}{6 \times 0.1}
=29.2= 29.2 mL.
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