A) 48 g
B) 70 g
C) 49 g
D) 35 g
Correct Answer: B
Solution :
The neutralization of NaOH by \[{{H}_{2}}S{{O}_{4}}\]takes place as follows \[{{H}_{2}}S{{O}_{4}}+2NaOH\xrightarrow{{}}N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O\] For complete neutralization Equivalents of acid = equivalents of base Equivalents of NaOH = moles \[\times \]acidity \[=1\times 1=1\] Equivalents of \[{{H}_{2}}S{{O}_{4}}=\frac{x}{98}\times 2=\frac{x}{49}\] (Mol. mass of \[{{H}_{2}}S{{O}_{4}}=98\]) Putting the values \[1\times 1=\frac{x}{49}\] \[\Rightarrow \] \[x=49\,g\] but \[{{H}_{2}}S{{O}_{4}}\]is 70% let y g 70% \[{{H}_{2}}S{{O}_{4}}\]is required \[\frac{70}{100}\times y=49\] \[\Rightarrow \] \[y=70\,g\]
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