JEE Main 2017 — Some Basic Concepts Of Chemistry Question with Solution
From: JEE Main 2017 (Online) 8th April Morning Slot
Question
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)
(Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)
Choose an option
Show full solutionCorrect option: A
Correct answer
A0.2 M
Step-by-step explanation
3 NaOH (aq.) + FeCl3(aq) Fe(OH)3(s)+ 3 NaCl(aq).
Moles of Fe(OH)3 = = 2 102
1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3
2 102 moles of Fe(OH)3 will obtain from
= 0.02 mole of FeCl3
Molarity of FeCl3 = = = 0.2 M
Moles of Fe(OH)3 = = 2 102
1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3
2 102 moles of Fe(OH)3 will obtain from
= 0.02 mole of FeCl3
Molarity of FeCl3 = = = 0.2 M
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Some Basic Concepts Of Chemistry chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2017, covering the Some Basic Concepts Of Chemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.