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Some Basic Concepts of Chemistry question

2021 · 17 Mar · Shift 1 · Q16
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Some Basic Concepts of Chemistry question

2021 · 17 Mar · Shift 1 · Q16

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
15 mL of aqueous solution of Fe2+Fe^{2+}Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr2OCr_2OCr2​O 72−_7^{2 - }72−​. The molarity of the Fe2+Fe^{2+}Fe2+ solution is ‾×\underline{\hspace{2cm}}\times​× 10-2 M. (Round off to the Nearest Integer).
Numerical answer
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Correct answer: 24

  1. Write the balanced redox reaction in acidic medium

In acidic medium,

Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2OCr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2OCr2​O72−​+14H++6Fe2+→2Cr3++6Fe3++7H2​O

So, the mole ratio is:

1 mol Cr2O72−:6 mol Fe2+1\text{ mol } Cr_2O_7^{2-} : 6\text{ mol } Fe^{2+}1 mol Cr2​O72−​:6 mol Fe2+

  1. Calculate moles of Cr2O72−Cr_2O_7^{2-}Cr2​O72−​ used

Given:

  • Volume of Cr2O72−Cr_2O_7^{2-}Cr2​O72−​ solution =20 mL=0.020 L= 20\,\text{mL} = 0.020\,\text{L}=20mL=0.020L
  • Molarity =0.03 M= 0.03\,\text{M}=0.03M

moles of Cr2O72−=M×V=0.03×0.020=6.0×10−4\text{moles of } Cr_2O_7^{2-} = M \times V = 0.03 \times 0.020 = 6.0 \times 10^{-4}moles of Cr2​O72−​=M×V=0.03×0.020=6.0×10−4

  1. Use stoichiometry to find moles of Fe2+Fe^{2+}Fe2+

From the balanced equation,

1 mol Cr2O72− reacts with 6 mol Fe2+1\text{ mol } Cr_2O_7^{2-} \text{ reacts with } 6\text{ mol } Fe^{2+}1 mol Cr2​O72−​ reacts with 6 mol Fe2+

Therefore,

moles of Fe2+=6×6.0×10−4=3.6×10−3\text{moles of } Fe^{2+} = 6 \times 6.0\times 10^{-4} = 3.6\times 10^{-3}moles of Fe2+=6×6.0×10−4=3.6×10−3

  1. Calculate molarity of Fe2+Fe^{2+}Fe2+ solution

Given volume of Fe2+Fe^{2+}Fe2+ solution:

15 mL=0.015 L15\,\text{mL} = 0.015\,\text{L}15mL=0.015L

So,

M=molesvolume=3.6×10−30.015=0.24 MM = \frac{\text{moles}}{\text{volume}} = \frac{3.6\times 10^{-3}}{0.015} = 0.24\,\text{M}M=volumemoles​=0.0153.6×10−3​=0.24M

  1. Express in the required form

We need:

0.24 M=24×10−2 M0.24\,\text{M} = 24 \times 10^{-2}\,\text{M}0.24M=24×10−2M

So the required integer is:

24\boxed{24}24​

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