A) 20.73%
B) 43.80%
C) 79.27%
D) 60.74%
Correct Answer: A
Solution :
\[{{H}_{2}}{{S}_{2}}{{O}_{7}}+{{H}_{2}}O\xrightarrow{{}}2{{H}_{2}}S{{O}_{4}}\] \[S{{O}_{3}}\]of\[{{H}_{2}}{{S}_{2}}{{O}_{7}}\]is converted into\[{{H}_{2}}S{{O}_{4}}\], hence\[S{{O}_{3}}\]acts also as a dibasic acid. Eq. wt. \[(S{{O}_{3}})=\frac{M}{2}=40\] Let\[{{H}_{2}}S{{O}_{4}}\]in fuming\[{{H}_{2}}S{{O}_{4}}=x\,\,g\] \[S{{O}_{3}}=(1-x)g\] Equivalent of\[{{H}_{2}}S{{O}_{4}}=\frac{x}{49}\] \[S{{O}_{3}}=\frac{1-x}{40}\] Equivalent of\[NaOH\]used\[=\frac{26.7\times 0.8}{1000}\] \[=0.02136\] \[\frac{x}{49}+\frac{1-x}{40}\,=0.02136\] \[x=0.7927g\,\,{{H}_{2}}S{{O}_{4}}\]in\[1g\] oleum. Percentage of\[{{H}_{2}}S{{O}_{4}}=79.27%\] \[S{{O}_{3}}=20.73%\]
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