The equilibrium constant for mutarotation α-D Glucose ͡

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 Multiple Choice QuestionsMultiple Choice Questions

601.

Assertion : PCl5, is covalent in gaseous and  liquid states but ionic in solid state.

Reason : PCl, in solid state consists of tetrahedral PCl4+cation and octahedral PCl6- anion.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


602.

Solubility product of a salt AB is 1 x 10-8 M in a solution in which the concentration of A+ ions is 10-3 M .The salt , will precipitate when the concentration of B- ions is kept :

  • between 10-8 M to 10-7 M

  • between 10-7 M to 10-8 M

  • > 10-5 M

  • < 10-8 M


603.

The pH of 10-8 M HCl solution is :

  • 8

  • more than 8

  • between 6 and 7

  • slightly more than 7


604.

For a reversible reaction ;

X (g) + 3Y (g)  2Z (g) ; H = -40 kJ , the standard entropies of X , Y and Z are 60 , 40 and 50JK mol-1 respectively .The temperature at which the above reaction attains equilibrium is about :

  • 400 K

  • 500 K

  • 273K

  • 373K


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605.

The dissociation equilibrium of a gas AB2 can be represented as

2AB2(g)  2 AB(g) + B2(g)

The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant KP and total pressure p is

  • (2Kp/p)1/2

  • Kp/p

  • 2Kp/p

  • (2kp/p)1/3


606.

pH of a 0.01 M solution (Ka = 6.6 × 10-4)

  • 7.6

  • 8

  • 2.6

  • 5


607.

The pH of a 0.02 M solution of HCl is :

  • 2.2

  • 2.0

  • 0.3

  • 1.7


608.

The solubility product of Mg(OH)2 at 25°C is 1.4 x 10-11 .What is the solubility of Mg(OH)2 in g/L ?

  • 0.0047 g/L

  • 0.047 g/L

  • 0.0087 g/L

  • 0.087 g/L


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609.

A vessel of one litre capacity containing 1 mole of SO3 is heated till a state of equilibrium is attained.

2SO3 (g)  2SO2 (g) + O2 (g)

At equilibrium, 0.6 moles of SO2 had formed. The value of equilibrium constant is

  • 0.18

  • 0.36

  • 0.45

  • 0.68


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610.

The equilibrium constant for mutarotation α-D Glucose  β-D Glucose is 1.8. What percentage of α form remains at equilibrium?

  • 35.7

  • 64.3

  • 55.6

  • 44.4


A.

35.7

InitialAt equilibrium α-D glucose11 - α  β-D glucose0αK = α1 - α

Solving α = 0.642

1 - α = 0.358

Percent of α-D glucose remaining at equilibrium = 35.8%


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