A) 50
B) 78.4
C) 80
D) 39.2
Correct Answer: A
Solution :
Eq. of \[KMn{{O}_{4}}\] used\[=\frac{50\times 1}{1000\times 10}=0.005\] \[\therefore \]Eq of FAS reacted =0.005 \[\therefore \]weight of FAS needed =0.005 x 392 =1.96 g Thus, percentage purity of FAS is 50%You need to login to perform this action.
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