Moles of Fe(OH)3 =1072.14=2×10−2mol
Moles FeCl3=100ml×10−3×M
According to stoichiometry, mole of FeCl3=MoleofFe(OH)3
100 × 10−3×M=2×10−2
M=1002×10−2×1000=0.2M
JEE Main 2017 — Chemistry Physical Chemistry
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 the molar mass ofFe=56gmol−1 and molar mass of Cl=35.5gmol−1)
Held on 8 Apr 2017 · Verified 6 Jul 2026.
0.3M
0.2M
0.6M
1.8M
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