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1. Some basic concept of chemistry — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry1. Some basic concept of chemistry1 Mark
MCQ
Excess of $NaOH\, (aq)$ was added to $100\, mL$ of $FeCl_3\, (aq)$ resulting into $2.14\, g$ of $Fe(OH)_3$ .The molarity of $FeCl_3\, (aq)$ is  (Given molar mass of $Fe=56\, g\, mol^{-1}$ and molar mass of .............. $\mathrm{M}$ ($Cl=35.5\, g\, mol^{-1}$)
  • ✓
    $0.2$
  • B
    $0.3$
  • C
    $0.6$
  • D
    $1.8$
✓

Answer

Correct option: A.
$0.2$
a
$\underset{\begin{smallmatrix} 
 \lim iting \\ 
 reagent 
\end{smallmatrix}}{\mathop{FeC{{l}_{3}}(aq.)}}\,+\underset{\begin{smallmatrix} 
 (Excess\,amount) \\ 
 not\,behave\,as 
 \\ 
 \lim iting\,reagent 
\end{smallmatrix}}{\mathop{3NaOH(aq.)}}\,\to $ $Fe{(OH)_3}(s) + 3NaCl(aq.)$

Moles of $Fe(OH)_3$ $ = \frac{{weight\,in\,g}}{{M.\,weight\,of\,Fe{{(OH)}_3}}}$

$ = \frac{{2.14\,g}}{{107\,g/mol}} = 0.02\,mol.$

$1.0$ mole of $Fe(OH)_3$ is obtained from $= 1.0$ mole of $FeCl_3$

$0.02$ mole of $Fe(OH)_3$ is obtained from $0.02$ mole of $FeCl_3$

Molarity

$ = \frac{{No.\,of\,moles}}{{Volume\,in\,L}} = \frac{{0.02\,mole}}{{0.1\,L}} = 0.2\,M$

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