Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

41.

Gold sol is not

  • a macromolecular colloid

  • a lyophobic colloid

  • a multimolecular colloid

  • negatively charged colloid


42.

For hydrogen-oxygen fuel cell at 1 atm and 298 K

H2 (g) + 12O2 (g) → H2O (l) ; G° = -240 kJ

E° for the cell is approximately, (Given F = 96500 C)

  • 2.48 V

  • 1.25 V

  • 2.5 V

  • 1.26 V


43.

The correct statement is

  • The earlier members of lanthanoid series resemble calcium in their chemical properties.

  • The extent of actinoid contraction is almost the same as lanthanoid contraction.

  • In general, lanthanoid and actinoids do not show variable oxidation states.

  • Ce4+ in aqueous solution is not known.


44.

For Freundlich isotherm a graph of log x/m is plotted against log p. The slope of the line and its y-axis intercept, respectively corresponds to

  • 1n , k

  • log 1n , k

  • 1n , log k

  • log 1n, log k


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45.

The IUPAC name of the complex ion formed when gold dissolves in aqua-regia is

  • tetrachloridoaurate (III)

  • tetrachloridoaurate (I)

  • tetrachloridoaurate (II)

  • dichloridoaurate (III)


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46.

A (g)  P (g) + Q (g) + R (g), follows first order kinetics with a half-life of 69.3 s at 500°C. Starting from the gas 'A' enclosed in a container at 500°C and at a pressure of 0.4 atm, the total pressure of the system after 230 s will be

  • 1.15 atm

  • 1.32 atm

  • 1.22 atm

  • 1.12 atm


D.

1.12 atm

At t = 0At time t A (g)0.4 atm(0.4 - x)  P(g)0x + Q(g) 0x + R(g)0x

pt = (0.4 - x) + x + x + x

   = 0.2 + 2x

or, x = pt - 0.42 pA = p0 - x = p0 - pt - 0.42                          = 2 × 0.4 - pt + 0.42                          = 1.2 - pt2From, k = 2.303t log p0pA  0.693t1/2  = 2.303230 log 0.4 × 21.2 - pt  0.69369.3 = 2.303230 log 0.81.2 - ptlog 0.81.2 - pt = 0.693 × 23069.3 × 2.303                            = 0.99870.81.2 - pt = antilog (0.9987)0.81.2 - pt  10 12 - 10 pt = 0.8 or pt = 1.12 atm


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47.

A crystalline sold XY3 has ccp arrangement for its element Y. X occupies

  • 66% of tetrahedral voids

  • 33% of tetrahedral voids

  • 66% of octahedral voids

  • 33% of octahedral voids


48.

Match the reactant in Column I with the reaction in Column II.

Column I Column II
A. Acetic acid i. Stephen
B. Sodium phenate ii. Friedel Crafts
C. Methyl cyanide iii. HVZ
D. Toulene iv. Kolbe's

  • A - iii; B - i; C - iv; D - ii

  • A - iv; B - ii; C - iii; D - i

  • A - ii; B - iii; C - i; D - iv

  • A - iii; B - iv; C - i; D - ii


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49.

(Reductive)Ozonolysis Y + Z

Y can be obtained by Etard's reaction, Z undergoes disproportionation reaction with concentrated alkali. X could be


50.

(ii) H3O+(i) CH3MgBr R (i) dil. NaOH

4-methylpent-3-en-2-one P is

  • propanone

  • ethanamine

  • ethanenitrile

  • ethanal


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