A) 70 g
B) 35 g
C) 30 g
D) 95 g
Correct Answer: A
Solution :
The neutralization of \[NaOH\] and \[{{H}_{2}}S{{O}_{4}}\] take place as follows \[2NaOH+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O\] 2 mole of \[NaOH\] is neutralized by 1 mole \[\therefore \]1 mole \[NaOH\] is neutralized by \[\frac{1}{2}\] mole \[NaOH\]. Given 70% \[{{H}_{2}}S{{O}_{4}}\] i.e., 70 g of \[{{H}_{2}}S{{O}_{4}}\] is dissolved in 100 mL acid Mol. mass of \[{{H}_{2}}S{{O}_{4}}\] \[=1\times 2+32\times 1+16\times 4=98g\] 1 mole of \[{{H}_{2}}S{{O}_{4}}=98g\] \[\frac{1}{2}\]mol \[{{H}_{2}}S{{O}_{4}}=49g\] 70 g \[{{H}_{2}}S{{O}_{4}}\] is present in 100 g acid \[\therefore \] 49 g will be present in \[\frac{100}{70}\times 49=70\text{ }g\] acid
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