A) 2.61 g
B) 5.22 g
C) 1.305 g
D) 2.78 g
Correct Answer: A
Solution :
Equivalent weight of \[FeS{{O}_{4}}.7{{H}_{2}}O=Mol.wt.=278\] 80 mL 0.125 (N) permanganate solution = (80 x 0.125) N solution = meq. of \[FeS{{O}_{4}}.7{{H}_{2}}O\] =10 Weight\[=\frac{10\times 278}{1000}=2.78g\] Weight of anhydrous \[F{{e}_{2}}{{(S{{O}_{4}})}_{3}}\] \[=5.39-2.78=2.61g\]You need to login to perform this action.
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