Question

25g of a mixture of NaOh and NaCl (.as impurity ) were dissolved in 500 cm of water. If 25cm^3 of this solution we're neutralized by 21cm^3 0.1mol/dm^3 of hydrochloric acid. Calculate the percentage of the NaCl impurity (NaOh=40,HCl=365,NaCl=58.5)

Answer 1

NaOH + HCl ==> NaCl + H2O
millimoles HCl used = mL x M = 21 cc x 0.1 = 2.1 or 0.0021 moles.
Since 1 mol NaOH = 1 mol HCl there were 0.0021 moles NaOH present and grams NaOH = mols x molar mass = 0.0021 moles x 40 g/mol = 0.084 g NaOH in the 25 mL sample of that solution. Then grams NaOH in the entire sample = 0.084 g x (500/25) = ?
Then %NaOH = (total grams NaOH/25 g sample)*100 = ?

Answer 2

oops. That's the percent NaOH I gave you.
100% - %NaOH = %NaCl.
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The other way, slightly shorter is to calculate grams NaOH as above. Then 25 total - g NaOH = grams NaCl. Then %NaCl = (grams NaCl/25)*100 = ?