JEE MAIN - Chemistry (2017 - 8th April Morning Slot - No. 24)
Excess of NaOH (aq) was added to 100 mL of FeCl3 (aq) resulting into 2.14 g of Fe(OH)3 . The molarity of FeCl3 (aq) is :
(Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)
(Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)
0.2 M
03 M
0.6 M
1.8 M
Explanation
3 NaOH (aq.) + FeCl3(aq) $$ \to $$ Fe(OH)3(s)+ 3 NaCl(aq).
Moles of Fe(OH)3 = $${{2.14} \over {107}}$$ = 2 $$ \times $$ 10$$-$$2
1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3
$$\therefore\,\,\,$$ 2 $$ \times $$ 10$$-$$2 moles of Fe(OH)3 will obtain from
= 0.02 mole of FeCl3
Molarity of FeCl3 = $${{No.of\,moles} \over {Volume\,in\,L}}$$ = $${{2 \times {{10}^{ - 2}}} \over {0.1}}$$ = 0.2 M
Moles of Fe(OH)3 = $${{2.14} \over {107}}$$ = 2 $$ \times $$ 10$$-$$2
1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3
$$\therefore\,\,\,$$ 2 $$ \times $$ 10$$-$$2 moles of Fe(OH)3 will obtain from
= 0.02 mole of FeCl3
Molarity of FeCl3 = $${{No.of\,moles} \over {Volume\,in\,L}}$$ = $${{2 \times {{10}^{ - 2}}} \over {0.1}}$$ = 0.2 M
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