Solution X contains Na2CO3 and NaHCO3, 20 mL of X when titrated

Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The number of radial nodes of 3s and 2p orbitals respectively are

  • 0, 2

  • 2, 0

  • 1, 2

  • 2, 1


2.

The basis of quantum mechanical model of an atom is

  • angular momentum of electron

  • quantum numbers

  • dual nature of electron

  • black body radiation


3.

The number of elements present in the fourth period is

  • 32

  • 8

  • 18

  • 2


4.

Identify the correct set.

  • Molecule Hybridisation of central atom Shape
    PCl5 dsp3 square pyramidal
  • [Ni(CN)4]2- sp3 tetrahedral
  • SF6 sp3d2 octahedral
  • IF3 dsp3 pyramidal

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5.

Which one of the following statements is correct?

  • Hybrid orbitals do not form σ bonds.

  • Lateral overlap of p-orbitals or p- and d-orbitals produces π-bonds.

  • The strength of bonds follows the order-

    σp-p < σs-sπp-p

  • s-orbitals do not form σ bonds.


6.

The degree of ionization of 0.10 M lactic acid is 4.0%

The value of Kc is

  • 1.66 × 10-5

  • 1.66 × 10-4

  • 1.66 × 10-3

  • 1.66 × 10-2


7.

The pH of a buffer solution made by mixing 25 mL of 0.02 M NH4OH and 25 mL of 0.2 M NH4Cl at 25° is pKb of NH4OH = 4.8)

  • 5.8

  • 8.2

  • 4.8

  • 3.8


8.

For which one of the following reactions, the entropy change is positive?

  • H2 (g) + 12O2 (g) → H2O (l)

  • Na+ (g) + Cl- (g) → NaCl (s)

  • NaCl (l) → NaCl (s)

  • H2O (l) → H2O (g)


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9.

Solution 'X' contains Na2CO3 and NaHCO3, 20 mL of X when titrated using methyl orange indicator consumed 60 mL of 0.1 M HCl solution. In another experiment, 20 mL of X solution when titrated using phenolphthalein, consumed 20 mL of 0.1 M HCl solution. The concentrations (in mol L-1) of Na2CO3 and NaHCO3 in X are respectively

  • 0.01, 0.02

  • 0.1, 0.1

  • 0.01, 0.01

  • 0.1, 0.01


B.

0.1, 0.1

For a titration of a basic solution of Na2CO3 and NaHCO3 against HCl, if phenolphthalein is used as indicator, the end product is indicated only for half neutralization of Na2CO3i.e. upto NaHCO3

Na2CO+ HCl → NaHCO+ NaCl

The remaining solution then contains the unreacted  NaHCO3 from this reaction plus the unreacted NaHCO3 originally in the solution. At the phenolphthalein end point, there is no reaction between HCl and NaHCO3.

From the equation,

Mole of HCl consumed = mol of Na2CO3

         10 mol of 0.1 M = 20 mol of 0.1 M

 The concentration of Na2CO3 in solution X = 0.1 M

For a quantity of Na2COexactly half volume of the HCl is used at the phenolphthalein
end point and the second half volume of the HCl is required for complete neutralization of Na2CO3 at methyl orange end point.

NaHCO3 + HCl → NaCl + CO2 + H2O

 The volume of HCl required to neutralize Na2COin original sample = 2 × 20 mL = 40 mL

If methyl orange is used, the end point is indicated when all the alkali is neutralized.

As 40 mL of 0.1 M HCl is consumed in complete neutralization of Na2CO3 at methyl ornage end point from the original sample would be reamining HCl

= (60 - 40) mL 

= 20 mL

Now, 

1 mol of NaHCO3 = 1 mol of HCl

 0.1 mol of NaHCO3 = 0.1 mol of HCl


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10.

A compound absorbs light in the wavelength region 490-500 nm. Its complementary colour is

  • red

  • blue

  • orange

  • blue-green


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