A) 10.84%
B) 14.64%
C) 16.66%
D) 18.52%
Correct Answer: D
Solution :
\[%\,S=\frac{32}{233}\times \frac{massof\,BaS{{O}_{4}}}{mass\,of\,subs\tan ce}\times 100\][Carius method] (mol. mass of \[BaS{{O}_{4}}=233\]) \[%\,\,S=\frac{32}{233}\times \frac{0.35}{0.2595}\times 100=18.52%\]You need to login to perform this action.
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