Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

0.1 M solution of CH3COOH should be diluted to how many times so that pH is doubled ?

  • 4.0 times

  • 5.55 x 104 times

  • 5.55 x 106 times

  • 10-2 times


B.

5.55 x 104 times

pH = 12[pKa - log1] = pKa2

pH' (twice of pH) = pKa

pKa12[pKa - logC]

logC = pKa = -logKa

C = Ka = 1.8 x 10-5 M

 Dilution = 1C = 11.8 × 5 = 5.55 x 104 times.


2.

For the process ,

X (g) + e-  X- (g) ; ΔH = x

and X- (g)  X (g) + e-ΔH = y

select correct alternate

  • Ionisation energy of x- (g) is y

  • Electron affinity of X (g) is x

  • Electron affinity of X (g) is - y

  • All of the above


D.

All of the above

X (g) + e-  X- (g) ; H = x = EA of X (g)

X- (g)  X (g) + e-H = y = IE of X- (g) 

= - EA of X (g)

Therefore , all options are correct .


3.

NO2 (brown colour gas) exists in equilibrium with N2O4 (colourless gas) as given by chemical equation.

2NO2(brown)  N2O4(colourless)

Mixture is slightly brown due to existence of NO2. If pressure is increased :

  • colour intensity is increased

  • colour intensity is decreased

  • colour intensity first increases and then decreases

  • No change in colour intensity


D.

No change in colour intensity

By Le-Chateliers principle, due to increase in pressure, equilibrium shifts in the direction of low pressure (in which number of mole decreases) i.e. , forward side.


4.

If the equilibrium constant of the reaction of weak acid HA with strong base is 109 then pH of 0.l M NaA is :

  • 5

  • 9

  • 7

  • 8


B.

9

HA + OH-  H2O + A-

K = [A-][HA][OH-]

HA  H+ + A-

Ka[H+][A-][HA]

kaK = [H+][OH-] = Kw

or    Ka = K2K = 10-14 × 109 = 10-5

pKa = 5

A- solution is alkaline due to hydrolysis

 pH = 7 + pKa2 + log C2

= 7 + 52 + log 0.12 = 9


5.

Mass of one atom is 6.66 x 10-23 g. Its percentage in an hydride is 95.24. Thus , hydride is :

  • MH

  • MH2

  • MH3

  • MH4


B.

MH2

Mass of one atom= 6.66 x 10-23 g

Mass of N0 atoms = 6.66 x 10-23 x 6.02 x 10-23 g = 40 g

Thus , atomic weight of the element = 40

Element     % % at.wt. Ratio
M 95.24 2.381 1
H 4.76 4.76 2

Thus , hydride is MH.


6.

The heat of formation of C12H22O11 (s) , CO2 (g) and (H2O) are - 530 , -94.3 and- 68.3 kcalmol-1 respectively. The amount of C12H22O11 to supply 2700 kcal of energy is :

  • 382.70 g

  • 832.74 g

  • 463.9 g

  • 684.0 g


D.

684.0 g

C12H22O11 (s) + 12O2 (g)  12CO2 (g) + 11H2O (l)

ΔHcomb° = [12ΔHf°(CO2) + 11ΔHf° (H2O)] - ΔHf° (C12H22O11)]

= - 1352.9 kcal

Thus, number of moles of C12H22O11 for getting 2700 kcal of heat.

27001352.9 = 2 mol

= 2 X 242 = 684.0 g


7.

1 mol of O2 and x mol of Ne in a 10 L flask at constant temperature exert a pressure of 10 atm. If partial pressure of O2 is 2 atm , moles of Ne in the mixture is :

  • 1

  • 2

  • 4

  • 3


C.

4

D.

3

pO2 = ptotal po2

po2 = 11 + x

po2 = 2 = 10 x 11 + x or 10 = 2 + 2x

x = 8/2 = 4 mol


8.

Spontaneous adsorption of gas on solid surface is an exothermic process because :

  • ΔH increases for system

  • ΔS increases for gas

  • ΔS decreases for gas

  • ΔG increases for gas


C.

ΔS decreases for gas

When a gas is adsorbed spontaneously on a solid surface, entropy decreases for gas.


9.

ΔHf° (C2H4) , ΔHf° (C2H6) are x1 and x2 kcalmol-1 respectively. Then heat of hydrogenation of C2H4 is :

  • x1 + x2

  • x1 - x2

  • -(x1 - x2)

  • x1 + 2x2


C.

-(x1 - x2)

C2H4 + H2 C2H6

ΔH = ΔHf° (C2H6) - ΔHf° (C2H4) - ΔHf° (H2)

= x2 - x1 - 0

= x2 - x1

= -(x1 - x2)


10.

A 1 L flask contains 32 g O2 gas at 27C. What mass of Omust be released to reduce the pressure in the flask to 12.315 atm?

  • 8 g

  • 16 g

  • 24 g

  • 32 g


B.

16 g

pV = nRT = wmRT

w = pVmRT = 12.315 × 1 × 320.0821 × 300 = 16 g

O2 to be released= 32 - 16 = 16 g