A piece of aluminium weighing 2.7 g is heated with 75.0 mL of sulphuric acid ( sp . Gr . 1.18 containing 24.7 % H_(2)SO_(4) be weight ) .Afterthe metal is carefully dissolved the solutions is dilutedto 400 mL . Calculate the molarity of the free H_(2)SO_(4)in the resulting solution .

A piece of aluminium weighing 2.7 g is heated with 75.0 mL of sulphuri

1.	A piece of aluminium weighing 2.7 g is heated with 75.0 mL of sulphuric acid ( sp . Gr . 1.18 containing 24.7 % H_(2)SO_(4) be weight ) .Afterthe metal is carefully dissolved the solutions is dilutedto 400 mL . Calculate the molarity of the free H_(2)SO_(4)in the resulting solution .
Answer» Solution :Normality of the given `H_(2)SO_(4) = 5.95 N ` ` :. ` m.e of 75 mL of `H_(2)SO_(4) = 5.95 xx75` ` = 446.25 ` Equivalent of Al ` = (2.7)/9 = 0.3 "" …(Eqn .4)` `{ " EQ .wt. of Al " = ("al .wt")/("valency") = 27/3 = 9 }` ` :." m.e of Al " = 0.3 xx 1000 = 300 "" ...(Eqn .3)` Since 300 m.e of Al will react with 300 m.eof `H_(2)SO_(4)` m.e of FREE `H_(2)SO_(4)`= TOTAL m.e of `H_(2)SO_(4)-300` `= 446.25 - 300` `= 146.25` Now the free `H_(2)SO_(4)` is DILUTED to 400 mL and we knw that the m.e of `H_(2)SO_(4)` does not change on dilution . ` :. ` normality of the diluted free `H_(2)SO_(4)`in the resulting solution ` = ( " m.e of free" H_(2)SO_(4))/(" volume (mL)")` ` = (146.25)/400 = 0.366` N . ` :. "molarity " = (0.366)/2 = 0.183 M"" ...(Eqn . 6i)`( basicity of `H_(2)SO_(4) = 2`)