A) 84.8%
B) 8.48%
C) 15.2%
D) 42.4%
Correct Answer: A
Solution :
[a] Let the amount of \[N{{a}_{2}}C{{O}_{3}}\] present in the mixture be x g. \[N{{a}_{2}}S{{O}_{4}}\] will not react with\[{{H}_{2}}S{{O}_{4}}\]. Then \[\frac{x}{53}~=\frac{20\times 0.1\times 10}{1000}\therefore x=1.60g\] \[\therefore \] Percentage of \[N{{a}_{2}}C{{\text{O}}_{3}}\] \[=\frac{1.06\times 100}{1.25}=84%\]%You need to login to perform this action.
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