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प्रश्न
Explain why dilute hydrochloric acid cannot be concentrated by boiling beyond 22.2%.
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उत्तर
Distillation slowly increases the strength of a weak hydrochloric acid solution until it has 22.2% HCl by weight and boils at 110°C. Once this level is reached, boiling cannot be used to make the acid stronger anymore. When this concentration is attained, boiling [constant boiling mixture or azeotrope] cannot be used to enhance the acid concentration any further.
Water and HCl are the main components of vapours that form before 110°C, although HCl molecules make up the majority of vapours that form above this temperature.
संबंधित प्रश्न
State one relevant observation for given reactions:
Action of dilute Hydrochloric acid on iron (II) sulfide.
Give a chemical test to distinguish between the given pairs of chemicals:
Sodium chloride solution and Sodium nitrate solution
How will the action of dilute hydrochloric acid enable you to distinguish between sodium carbonate and sodium sulphite?
State which component is the oxidizing agent in aqua regia.
Name the following:
Drying agent used to dry hydrogen chloride.
Fill in the blank:
On addition of silver nitrate to hydrochloric acid ___________ precipitate is formed which is soluble in ____________
Write balanced equation for the following reaction:
Copper oxide and dilute hydrochloric acid.
Match the following:
| Column A | Column B |
| 1. A substance that turns moist starch iodide paper blue. | A. Ammonium sulphate |
| 2. A compound which release a reddish brown gas on reaction with concentrated sulphuric acid and copper turnings. | B. Lead carbonate |
| 3. A solution of this compound gives dirty green precipitate with sodium hydroxide. | C. Chlorine |
| 4. A compound which on heating with sodium hydroxide produces a gas which forms dense white fumes with hydrogen chloride. | D. Copper nitrate |
| 5. A white solid which gives a yellow residue on heating | E. Ferrous sulphate |
Select from the list given (a to e) one substances in each case which matches the description given in parts (i) to (v). (Note : Each substance is used only one in the answer)
(a) Nitroso Iron (II) Sulphate
(b) Iron (III) chloride
(c) Chromium sulphate
(d) Lead (II) chloride
(e) Sodium chloride
(i) A compound which is deliquescent
(ii) A compound which is insoluble in cold water, but soluble in hot water
(iii) The compound responsible for the brown ring during the brown ring test of nitrate iron
(iv) A compound whose aqueous solution is neutral in nature
(v) The compound which is responsible for the green colouration when sulphur dioxide is passed through acidified potassium dichromate solution
State your observation when:
Decomposition of bicarbonates by dil. H2SO4
2NaHCO3 + H2SO4 → Na2SO4 + 2H2O + 2CO2
2KHCO3 + H2SO4 → K2SO4 + 2H2O + 2CO2
