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`50 mL` of `1 M HCl, 100 mL` of `0.5 M HNO_(3)`, and `x mL` of `5 M H_(2)SO_(4)` are mixed together and the total volume is made upto `1.0 L` with water. `100 mL` of this solution exactly neutralises `10 mL` of `M//3 Al_(2) (CO_(3))_(3)`. Calculate the value of `x`. |
Answer» Correct Answer - A Total mEq of acid `= 50 xx 1 + 100 xx 0.5 + x xx 5 xx 2` (`n` factor ) `= (100 + 10x) mEq` `= ((100 + 10 x))/(100 mL) N` `N_(1) V_(1)` (Acid) `= N_(1) V_(1) [Al_(2) (CO_(3)^(2-))_(3)]` (Total charge = 6) `(n = 6)` `:. ((100 + 10 x))/(100 mL) N xx 100 mL = 10 mL xx (1)/(3) xx 6` `(100 + 10x) = 200` `:. x = 10 mL` |
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