Chemistry - Stoichiometry Question with Solution | TestHub
ChemistryStoichiometryProblems Based on Mixtures, Percentage PurityHard2 minQB
ChemistryHardsingle choice
Passage / Comprehension
{"root":{"type":"root","format":"","indent":0,"version":1,"children":[{"type":"paragraph","format":"","indent":0,"version":1,"children":[{"mode":"normal","text":"Two grams of a moist sample of a mixture of sodium chloride and sodium chlorate was dissolved in water and made up to 250 ml. Twenty-five millilitres of this solution was treated with ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{SO_{2}}"},{"mode":"normal","text":" to reduce the chlorate to chloride and excess ","type":"text","style":"","detail":0,"format":0,"version":1},{"type":"equation","format":"latex","inline":true,"version":1,"equation":"\\mathrm{SO_{2}}"},{"mode":"normal","text":" was removed by boiling. The total chloride was precipitated as silver chloride. The weight of the precipitate was 0.3444 g. In another experiment, 25 ml of the original solution was heated with 40 ml of 0.2 N solution of ferrous sulphate and unreacted ferrous sulphate required 10 ml of 0.08 N solution of an oxidizing agent for complete oxidation. (Na = 23, Ag = 108, Cl = 35.5)","type":"text","style":"","detail":0,"format":0,"version":1}],"direction":"ltr","textFormat":0}],"direction":"ltr"}}
What is the molar ratio of the chlorate to chloride in the given mixture?
Options:
Answer:
B
Solution:
= 0.3444/143.5 = 2.4 mmol, so total (after reduction) in 25 ml = 2.4 mmol, i.e. 24 mmol in 250 ml. Chlorate consumed = 40 × 0.2 - 10 × 0.08 = 8 - 0.8 = 7.2 meq per 25 ml; with ( = 6), chlorate = 1.2 mmol per 25 ml, i.e. 12 mmol in 250 ml. = 24 - 12 = 12 mmol. Ratio chlorate : chloride = 12 : 12 = 1:1, matching (b).
Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Problems Based on Mixtures, Percentage Purity
⏱ 2mℹ️ Source: QB
Doubts & Discussion
Loading discussions...
