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(a) What is the normality of a 96 per cent solution of H_(2)SO_(4) of specific gravity 1.84 ? (b) How many mL of 96 per cent sulphuric acid solution is necessary to prepare one litre 0.1 N H_(2)SO_(4) ? ( c) To what volume should 10 mL of 96 per cent H_(2)SO_(4) be diluted to prepure 2 N solution ?

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a> of `H_(2)SO_(4)` solution <br/> `="<a href="https://interviewquestions.tuteehub.com/tag/vol-723961" style="font-weight:bold;" target="_blank" title="Click to know more about VOL">VOL</a>."xx"Density"` <br/> `=1000xx1.84=1840 g` <br/> Mass of `H_(2)SO_(4)` present in one litre 96% `H_(2)SO_(4)` solution <br/> `= (96)/(100)xx1840=1766.4 g` <br/> Strength of `H_(2)SO_(4)` solution `=1766.4 g//L` <br/> (a) Normality `=("Strength")/("Eq. mass")=(1766.4)/(49)=36.05 N` <br/> (b) Let the volume taken be `V_(1)` mL <br/> Applying `"" N_(1)V_(1)=N_(2)V_(2)` <br/> `N_(1)=36.05 N, V_(1) = ?, N_(2)=(N)/(10), V_(2)=1000` mL <br/> `36.05xxV_(1)xx1000` <br/> So, `V_(1)=(1000)/(36.05xx10)=2.77` mL <br/> i.e., 2.77 mL of `H_(2)SO_(4)` is <a href="https://interviewquestions.tuteehub.com/tag/diluted-7675406" style="font-weight:bold;" target="_blank" title="Click to know more about DILUTED">DILUTED</a> to one litre. <br/> (c ) ` underset("Before dilution")(N_(B)V_(B))= underset("After dilution ")(N_(A)V_(A))` <br/> `10xx36.05=V_(A)xx2` <br/> `V_(A)=180.25` mL <br/> i.e., 10 mL of given `H_(2)SO_(4)` is diluted to 180.25 mL.</body></html>


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