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 Cl =35.5 g mol -1 )

Question

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 Cl =35.5 g mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/a4bc8e4cfa4fc1569d0be79da513a7b5-.png" /> Moles of Fe ( OH )3=(2.14/107) =2 × 10-2 mol moles of FeCl 3 =2 × 10-2 mol M=(2 × 10-2/100) × 1000=0.2 M

Q. Excess of $N a O H (a q)$ was added to $100 m L$ of $F e C l_{3} (a q)$ resulting into $2.14 g$ of $F e (O H)_{3}$ . The molarity of $F e C l_{3} (a q)$ is
(Given molar mass of $F e = 56 g m o l^{- 1}$ and molar mass of $Cl = 35.5 g m o l^{- 1}$ )

Moles of $F e (O H)_{3} = \frac{2.14}{107}$
$= 2 \times 1 0^{- 2} m o l$
moles of $F e C l_{3}$
$= 2 \times 1 0^{- 2} m o l$
$M = \frac{2 \times 1 0 ^{- 2}}{100} \times 1000 = 0.2 M$

Q. Excess of NaOH(aq) was added to 100mL of FeCl3​(aq) resulting into 2.14g of Fe(OH)3​. The molarity of FeCl3​(aq) is (Given molar mass of Fe=56gmol−1 and molar mass of Cl=35.5gmol−1 )

Moles of Fe(OH)3​=1072.14​ =2×10−2mol moles of FeCl3​ =2×10−2mol M=1002×10−2​×1000=0.2M

Q. Excess of $N a O H (a q)$ was added to $100 m L$ of $F e C l_{3} (a q)$ resulting into $2.14 g$ of $F e (O H)_{3}$ . The molarity of $F e C l_{3} (a q)$ is
(Given molar mass of $F e = 56 g m o l^{- 1}$ and molar mass of $Cl = 35.5 g m o l^{- 1}$ )

Moles of $F e (O H)_{3} = \frac{2.14}{107}$
$= 2 \times 1 0^{- 2} m o l$
moles of $F e C l_{3}$
$= 2 \times 1 0^{- 2} m o l$
$M = \frac{2 \times 1 0 ^{- 2}}{100} \times 1000 = 0.2 M$