A) 50
B) 78.4
C) 80
D) 39.2
Correct Answer: A
Solution :
Equation of KMn04 used \[=\frac{50\times 1}{100\times 10}=0.005\] \[\therefore \] Equation of Mohr's salt reacted \[=0.005\] \[\therefore \] Weight of Mohr's salt needed \[=0.005\times 392=1.96g\] Thus, percentage purity of Mohr's salt \[=\frac{1.96}{3.92}\times 100=50%\]You need to login to perform this action.
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