How many grams of concentrated nitric acid solution should be use

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 Multiple Choice QuestionsMultiple Choice Questions

601.

How many litres of ammonia gas at S.T.P. would 1 be needed to prepare 100 ml of 2.5 M ammonium hydroxide solution?

  • 5.6 L

  • 0.056 L

  • 11.2 L

  • 0.56 L


602.

If 0.189 g of a chlorine containing organic compound gave 0.287 g of silver chloride, then the percentage of chlorine in the organic compound is

  • 35.47

  • 35.57

  • 37.57

  • 45.37


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603.

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0M HNO3 ? The concentrated acid is 70% HNO3.

  • 45.0 g conc. HNO3

  • 90.0 g con. HNO3

  • 70.0 g conc. HNO3

  • 54.0 g conc. HNO3


A.

45.0 g conc. HNO3

Given, molairty of solution = 2Volume of solution = 250 mL =2501000=14LMolar mass of HNO3= 1 +14+ 3 ×16 = 63 g mol-1therefore, Molarity= weight of HNO3mass of HNO3× volume of solution (L)therefore,weight of HNO3 = molarity × mol. mass × volume(L)                              = 2 ×63×14g = 31.5 gIt is the weight of 100% HNO3.But the given acid is 70% HNO3therefore, Its weight = 31.5 × 10070g = 45 g


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604.

The system that contains the maximum number of atoms is

  • 4.25 g of NH3

  • 8 g of O2

  • 2 g of H2

  • 4 g of He


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605.

The weight of oxalic acid that will be required to prepare a 1000 ml (N/20) solution is:

  • 126/100 g

  • 63/40 g

  • 63/20 g

  • 126/20 g


 Multiple Choice QuestionsLong Answer Type

606.

In 'x' mL 0.3 N HCl, the addition of 200 mL distilled water or addition of 100 mL 0.1 N NaOH, gives the same final acid strength. Determine 'x'.


607.

Compound A treated with NaNH2 followed by CH3CH2Br gave compound B. Partial hydrogenation of compound B produced compound C, which on ozonolysis gave a carbonyl compound D, (C2H5O). Compound D did not respond to the iodoform test with I2/KI and NaOH. Find out the structures of A, B, C and D.


 Multiple Choice QuestionsMultiple Choice Questions

608.

The volume of ethyl alcohol (density 1.15 g/cc) that has to be added to prepare 100 cc of 0.5 M ethyl alcohol solution in water is

  • 1.15 cc

  • 2 cc

  • 2.15 cc

  • 2.30 cc


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609.

2 g of metal carbonate is neutralised completely by 100 mL of 0.1 N HCI. The equivalent weight of metal carbonate is:

  • 50

  • 100

  • 150

  • 200


610.

An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitate would be

  • 0.001

  • 0.002

  • 0.003

  • 0.01


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