Verified Solution • 2025–26 Session moderate

QJEEM17Ch7918: JEE Main 2017 Chemistry - Class 11 JEEM17Ch7918 Solution: Excess of NaOH (aq) was added to 100 mL of FeCl 3 (aq)…

Excess of NaOH (aq) was added to 100 mL of FeCl3 (aq) resulting into 2.14 g of Fe(OH)3 . The molarity of FeCl3 (aq) is :

(Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)

Multiple Choice Options

  • A 0.2 M
  • B 03 M
  • C 0.6 M
  • D 1.8 M

Step-by-Step Verified Solution

3 NaOH (aq.) + FeCl3(aq) → Fe(OH)3(s)+ 3 NaCl(aq).

Moles of Fe(OH)3 = 2.14107 = 2 × 10−2

1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3

∴ 2 × 10−2 moles of Fe(OH)3 will obtain from

= 0.02 mole of FeCl3

Molarity of FeCl3 = No.ofmolesVolumeinL = 2×10−20.1 = 0.2 M

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