Excess of NaOH (a q) was added to 100 m L 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 (a q) was added to 100 m L 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) 0.2 M (B) 0.3 M (C) 0.6 M (D) 1.8 M

Tardigrade Admin · Accepted Answer

3 NaOH + FeCl 3 arrow Fe ( CH )3+ NaCl 100 ml 2.14 gm m =? Moles of Fe ( CH 3)=(2.14/107)=2 × 10-2 mol moles 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 $F e C l_{3}$ (aq) 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 $Cl = 35 \cdot 5 g m o l^{- 1})$

$0.2 M$

$0.3 M$

$0.6 M$

$1.8 M$

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

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)

0.2M

0.3M

0.6M

1.8M