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Some Basic Concepts of Chemistry question

2017 · 8 Apr · Shift 1 · Q24
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Some Basic Concepts of Chemistry question

2017 · 8 Apr · Shift 1 · Q24

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
Excess of NaOHNaOHNaOH (aq) was added to 100 mL of FeCl3FeCl_3FeCl3​ (aq) resulting into 2.14 g of Fe(OH)3Fe(OH)_3Fe(OH)3​ . The molarity of FeCl3FeCl_3FeCl3​ (aq) is : (Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)
  1. A
    0.2 M
  2. B
    03 M
  3. C
    0.6 M
  4. D
    1.8 M
View written solutionFree

Correct answer: A

  1. Write the reaction

When excess NaOHNaOHNaOH is added to FeCl3FeCl_3FeCl3​, the reaction is:

FeCl3+3NaOH→Fe(OH)3↓+3NaClFeCl_3 + 3NaOH \rightarrow Fe(OH)_3 \downarrow + 3NaClFeCl3​+3NaOH→Fe(OH)3​↓+3NaCl

From the equation,

1 mol FeCl3→1 mol Fe(OH)31\text{ mol } FeCl_3 \rightarrow 1\text{ mol } Fe(OH)_31 mol FeCl3​→1 mol Fe(OH)3​

So, moles of FeCl3FeCl_3FeCl3​ initially present = moles of Fe(OH)3Fe(OH)_3Fe(OH)3​ formed.

  1. Calculate molar mass of Fe(OH)3Fe(OH)_3Fe(OH)3​

Given:

  • Molar mass of Fe=56Fe = 56Fe=56 g mol−1^{-1}−1
  • Molar mass of O=16O = 16O=16 g mol−1^{-1}−1
  • Molar mass of H=1H = 1H=1 g mol−1^{-1}−1

M(Fe(OH)3)=56+3(16+1)=56+51=107 g mol−1M(Fe(OH)_3) = 56 + 3(16+1) = 56 + 51 = 107\text{ g mol}^{-1}M(Fe(OH)3​)=56+3(16+1)=56+51=107 g mol−1

  1. Find moles of Fe(OH)3Fe(OH)_3Fe(OH)3​ formed

Mass of precipitate = 2.142.142.14 g

n(Fe(OH)3)=2.14107=0.02 moln\big(Fe(OH)_3\big) = \frac{2.14}{107} = 0.02\text{ mol}n(Fe(OH)3​)=1072.14​=0.02 mol

Therefore,

n(FeCl3)=0.02 moln(FeCl_3) = 0.02\text{ mol}n(FeCl3​)=0.02 mol

  1. Calculate molarity of FeCl3FeCl_3FeCl3​ solution

Volume of solution = 100100100 mL = 0.10.10.1 L

M=nV=0.020.1=0.2 MM = \frac{n}{V} = \frac{0.02}{0.1} = 0.2\text{ M}M=Vn​=0.10.02​=0.2 M

  1. Check options
  • A: 0.2 M0.2\,M0.2M ✅
  • B: 0.3 M0.3\,M0.3M ❌
  • C: 0.6 M0.6\,M0.6M ❌
  • D: 1.8 M1.8\,M1.8M ❌

Hence, the correct answer is A.

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