SCIENCE
CHEMISTRY
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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0.25 mol dm-3
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0.3 mol dm-3
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0.5 mol dm-3
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1.0 mol dm-3
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Detailed explanation-1: -What Happened in the Titration: In the reaction between NaOH and HCl, 1 mole of NaOH reacts with 1 mole of HCl. 1 NaOH + 1 HCl → 1 NaCl + 1 H2O (Mole ratio of NaOH to HCl is 1:1) • The concentration of the NaOH was 0.1 M, or 0.1 moles/liter.
Detailed explanation-2: -The Na2CO3 content of the given sample is determined by titration against standard hydrochloric acid using methyl orange as indicator: Na2CO3 + 2HCl = 2NaCl + H2O + CO2. The equivalent weight of Na2CO3 is half its molecular weight (M). This means M/2 g of Na2CO3 is equivalent to 1000 ml of 1 N HCl.
Detailed explanation-3: -Concentration in mol/dm 3 = Concentration in mol/dm 3 = = 0.125 mol/dm 3 Relative formula mass of HCl = 1 + 35.5 = 36.5. Mass = relative formula mass × amount. Mass of HCl = 36.5 × 0.125. = 4.56 g. So concentration = 4.56 g/dm 3
Detailed explanation-4: -Amount of solute in mol = concentration in mol/dm 3 × volume in dm 3 Amount of sodium hydroxide = 0.100 × 0.0250. = 0.00250 mol. The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H 2O(l) So the mole ratio NaOH:HCl is 1:1. Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl. More items