The percentage (by weight) of sodium hydroxide in a 1.25 molal Na

Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The volume of 2 N H2SO4 solution is 0.1 dm3. The volume of its decinormal solution (in dm3) will be

  • 0.1

  • 0.2

  • 2

  • 1.7


2.

An element xAy emits 5 α and 4β particles to give 82B207. The number of protons and neutrons in A are respectively

  • 88, 227

  • 88, 139

  • 82, 227

  • 84, 139


3.

The degree of dissociation of a 0.01 M weak acid is 10-3. Its pOH is

  • 5

  • 3

  • 9

  • 11


4.

0.1 M solution of which of the following has almost unity degree of dissociation ?

  • Ammonium chloride

  • Potassium chloride

  • Sodium acetate

  • All of the above


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5.

The percentage (by weight) of sodium hydroxide in a 1.25 molal NaOH solution is

  • 4.76%

  • 1.25%

  • 5%

  • 40%


A.

4.76%

1.25 molal NaOH solution means, 1.25 moles of NaOH are present in 1000 gm of water.

 Weight of NaOH = 1.25 × 40 = 50 gm

Weight of solution = 1000 + 50 = 1050 gm

%(by weight) of NaOH = wt. of solutewt. of solution × 100

                                  = 501050 × 100

                                  = 4.76 %


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6.

If 20 mL of an acidic solution of pH 3 is diluted to 100 mL, the H+ ion concentration will be

  • 1 × 10-3 M

  • 2 × 10-3 M

  • 2 × 10-4 M

  • 0.02 × 10-4 


7.

At the same conditions of pressure, volume and temperature, work done is maximum for which gas if all gases have equal masses ?

  • NH3

  • N2

  • Cl2

  • H2S


8.

16 g of oxygen gas expands isothermally and reversibly at 300 K from 10 dm3 to 100 dm3. The work done is (in J)

  • zero

  • - 2875 J

  • + 2875 J

  • infinite


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9.

The calculated mass of 20Ca40 is 40.328 u. It releases 306.3 MeV energy in a nuclear process. Its isotopic mass is

  • 39.998

  • 40.6570

  • 0.3290

  • 2.85 × 104


10.

The maximum number of unpaired electrons is present in

  • Fe

  • Cu

  • Co

  • Ni


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