Excess of NaOH ( aq ) was added to 100mL of FeCl _ 3 (aq) resulting into 2 . 14g of Fe (OH) _ 3 . The molarity of FeCl _ 3 (aq) is: (Given the molar mass of Fe = 56g mol ^ - 1 and molar mass of Cl = 35 .5 g mol ^ - 1 )

Correct answer: B. 0 . 2M Moles of Fe ( OH ) _ 3 = (2 .14)/(107) = 2 × 10 ^ - 2 mol Moles FeCl _ 3 = 100ml × 10 ^ - 3 × M According to stoichiometry, mole of FeCl _ 3 = MoleofFe ( OH ) _ 3 100 × 10 ^ - 3 × M = 2 × 10 ^ - 2 M = 2 × 10 ^ - 2 100 × 1000 = 0 .2 M

JEE Main 20178 AprilOnline

Exam: JEE Main
Year: 2017
Date: 8 April
Mode: Online
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2017 · 8 April · Online · Chemistry · Some Basic Concepts of Chemistry

Excess of NaOH ( aq ) was added to 100mL of FeCl _ 3 (aq) resulting...

Excess of

NaOH ( aq )

was added to

100mL

{FeCl}_{3}

\left(aq\right)

resulting into

2 . 14g

of

Fe{\left(OH\right)}_{3}

. The molarity of

{FeCl}_{3}\left(aq\right)

is:<br /><br />(Given the molar mass of

Fe = 56g$

{mol}^{- 1} and molar mass of

Cl = 35 .5 g{mol}^{- 1}$)

Options

A $0 . 3M$
B $0 . 2M$
C $0 . 6M$
D $1 . 8M$

Correct answer

B. $0 . 2M$

Step-by-step solution

Moles of

Fe{\left(\text{ OH }\right)}

_{3} = \dfrac{2 .14}{107} = 2 \times {10}^{- 2}mol

<br /><br />Moles

{FeCl}_{3} =$

100ml \times {10}^{- 3} \times M

According to stoichiometry, mole of

{FeCl}_{3} = MoleofFe ( OH{)}_{3}

{\text{100 \times 10}}^{- 3} \times M = 2 \times {10}^{- 2}

$M = \dfrac{2 \times {10}^{- 2}}{100} \times 1000 = 0 .2 M

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