A bottle of commercial sulphuric acid (density = 1.787 gm/mL) is labelled 86 percent by weight. What is the molarity of the solution ? What volime of the acid is required to make it 1 litre of 0.2 M H_(2)SO_(4) ?

A bottle of commercial sulphuric acid (density = 1.787 gm/mL) is label

1.	A bottle of commercial sulphuric acid (density = 1.787 gm/mL) is labelled 86 percent by weight. What is the molarity of the solution ? What volime of the acid is required to make it 1 litre of 0.2 M H_(2)SO_(4) ?
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Step I. Molarity of solution <br/> Mass of `H_(2)SO_(4)=86 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>` , Mass of the solution = 100 g <br/> Volume of 100 g of the solution `= ("Mass")/("Density") = (100)/(1.787) = 55.96 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> = 0.05596 L` <br/> Molarity of solution (M) `= (("Mass of "H_(2)SO_(4))/("Molecular mass of "H_(2)SO_(4)))/("Volume of solution in mL"/(1000))=((86g))/(("<a href="https://interviewquestions.tuteehub.com/tag/98-342802" style="font-weight:bold;" target="_blank" title="Click to know more about 98">98</a> g mol"^(-1))xx("0.05596 L"))` <br/> `=15.68 "mol L"^(-1)=15.68 M`<br/> Step II. Volume of solution required <br/> `overset(("conc."))(M_(1)V_(1))-=overset(("dilute"))(N_(2)V_(2))` <br/> `15.68xxV_(1)=0.2xx1000,V_(1)=(0.2xx1000)/(15.68)=12.75mL`.</body></html>