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8 g of mixture of NaNO_(3) and Na_(2)CO_(3) are dissolved and made into 500 mL of solution. 50 mL of this solution neutralises completely 25 mL of N/5 HNO_(3). Calculate the precentage composition of the mixture. |
Answer» Solution :`Na_(2)CO_(3) + 2HNO_(3) rarr 2NaNO_(3) + H_(2)CO_(3)` Only sodium carbonate reacts with acid `(HNO_(3)) N_(1) V_(1) = N_(2) V_(2)` Normality of `Na_(2)CO_(3) " is " (0.2 xx 25)/(50) = 0.1 N` `:. 0.1 = ("wt of " Na_(2)CO_(3))/(53) xx (1000)/(500)` `:.wt " of " Na_(2)CO_(3) = (0.1 xx 53 xx 500)/(1000) = 2.65g` `:. wt " of " NaNO_(3) " is " = 8 - 2.65 = 5.35g` % of `NaNO_(3) = (5.35)/(8) xx 100 = 66.8%` % of `Na_(2)CO_(3) = (2.65)/(8) xx 100 = 33.2%` |
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