A piece of aluminium weighing 2.7 g is heated with 75 mL of H_(2)SO_(4) which has a density of 1.18"g mL"^(-1) and contains 24.7% by mass. When whole of the metal had dissolved, the solution was diluted to 400 mL. Calculate the molarity of free H_(2)SO_(4) in the resulting solution. Hence, molarity =(7.17)/(98)xx(1)/(400)xx1000=0.183M

A piece of aluminium weighing 2.7 g is heated with 75 mL of H_(2)SO_(4

1.	A piece of aluminium weighing 2.7 g is heated with 75 mL of H_(2)SO_(4) which has a density of 1.18"g mL"^(-1) and contains 24.7% by mass. When whole of the metal had dissolved, the solution was diluted to 400 mL. Calculate the molarity of free H_(2)SO_(4) in the resulting solution. Hence, molarity =(7.17)/(98)xx(1)/(400)xx1000=0.183M
Answer» Solution :`24.7%" of "H_(2)SO_(4)" by mass means 24.7 g of "H_(2)SO_(4)" are PRESENT in 100 g of the solution."` `"VOLUME of 100 g of "H_(2)SO_(4)" solution "=("Mass")/("Density")=(100g)/(1.18gmL^(-1))=84.7mL` Thus, `"84.7 mL of "H_(2)SO_(4)" soluiton contain "H_(2)SO_(4)=24.7g` `therefore "75 mL of "H_(2)SO_(4) " solution will contain "H_(2)SO_(4)=(24.7)/(84.7)xx75g=21.87g` The equaiton for the reaction is `underset(=54g)underset(2xx27g)(2Al)+underset(=294g)underset(3xx98g)(3H_(2)SO_(4))rarrAl_(2)(SO_(4))_(3)+3H_(2)` `"54 g Al react with "H_(2)SO_(4)=294g` `therefore "2.7 g Al will react with "H_(2)SO_(4)=(294)/(54)xx2.7g=14.7g` `thereforeH_(2)SO_(4)" left unreacted "=21.87-14.7=7.17g=(7.17)/(98)" moles"` This is present in 400 mLof the solution (after dilution).