The ratio of heats liberated at 298 K from the combustion of one

Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The statement that is not correct is

  • A furnace filled with haemalite is used to convert cast iron to wrought iron

  • Collectors enhance the wettability of mineral particles during froth flotation

  • In vapour phase refining, metal should form a volatile compound

  • Copper from its low grade ores is extracted by hydrometallurgy


2.

For Cr2O72- + 14 H+ + 6e- → 2Cr3+ + 7H2O; E° = 1.33 V. At [Cr2O72-] = 4.5 millimole, [Cr3+] = 15 millimole, E is 1.067 V. The pH of the solution is nearly equal to

  • 2

  • 3

  • 5

  • 4


3.

1.78 g of an optically active L-amino acid (A) is treated with NaNO3/ HCl at 0°C. 448 cm3 of nitrogen was at STP is evolved. A sample of protein has 0.25% of this amino acid by mass. The molar mass of the protein is

  • 36500 g mol-1

  • 34500 g mol-1

  • 35400 g mol-1

  • 35600 g mol-1


4.

10 g of a mixture of BaO and CaO requires 100 cm3 of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately (Given: molar mass of BaO = 158)

  • 52.6

  • 55.1

  • 44.9

  • 47.4


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5.

The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from kg of coke is (Assume coke to be 100% carbon). (Given : enthalpies of combustion of CO2, CO and H2 as 393.5 kJ, 285 kJ, 285 kJ respectively all at 298 K)

  • 0.79 : 1

  • 0.69 : 1

  • 0.86 : 1

  • 0.96 : 1


B.

0.69 : 1

Number of moles in 1 kg coke = 100012 = 83.33 mol

(i) For the combustion of 1 kg of coke

C + O2 → CO2H = 393.5 kJ

Heat liberated from 1 mole coke = 393.5 kJ

 Heat liberated from 83.33 mole coke = (393.5 × 83.33) kJ

(ii) For the burning of water gas

C + H2O → CO + H2

CO + H+ O2 → CO2 + H2O

H = 285 + 285 = 570 kJ

 Ratio of heat liberated from burning of water gas obtained from 1 kg of coke

H2 = (570 × 83.33) kJ

Therefore, required ratio = H1.H2

                                    = 393.5 × 570

                                   = 0.69 : 1


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6.

Impure copper containing Fe, Au, Ag as impurities is electrolytically refined. A current of 140 A for 482.5 s decreased the mass of the anode by 22.26 g and increased the mass of cathode by 22.011 g. Percentage of iron in impure copper is (Given molar mass Fe = 55.5 g mol-1, molar mass Cu = 63.54 g mol-1)

  • 0.95

  • 0.85

  • 0.96=7

  • 0.90


7.

25 cm3 of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

  • 0.064

  • 0.045

  • 0.015

  • 0.032


8.

The statement that is not correct is

  • Angular quantum number signifies the shape of the orbital

  • Energies of stationary states in hydrogen like atoms is inversely proportional to the square of the principal quantum number

  • Total number of nodes for 3s-orbital is three

  • The radius of the first orbit of He is half that of the first orbit of hydrogen atom


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9.

For the equilibrium,

CaCO3 (s)  CaO (s) + CO2 (g) ; 

Kp = 1.64 atm at 1000 K

50 g of CaCO3 in a 10 L closed vessel is heated to 1000 K. Percentage of CaCO3 that remains unreacted at equilibrium is (Given, R = 0.082 L atm K-1 mol-1).

  • 40

  • 50

  • 60

  • 20


10.

Density of carbon monoxide is maximum at

  • 2 atm and 600 K

  • 0.5 atm and 273 K

  • 6 atm and 1092 K

  • 4 atm and 500 K


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