﻿ NEET Chemistry Solved Question Paper 2007 | Previous Year Papers | Zigya

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# NEET Chemistry Solved Question Paper 2007

#### Multiple Choice Questions

1.

What is the frequency of a X-ray photon whose momentum is 1.1 x 10-23 kg-ms-2 ?

• 5 x 1016 Hz

• 0.5 x 1027 Hz

• 0.5 x 1018 Hz

• 5 x 1018 Hz

D.

5 x 1018 Hz

E = mc2

E = hv

mc2 = hv

$\mathrm{\nu }$ = $\frac{{\mathrm{mc}}^{2}}{\mathrm{h}}$ =

= 5  x 1018 Hz

2.

The bond energy is the energy required to :

• dissociate one mole of the substance

• dissociate bond in 1 kg of the substance

• break one mole of similar bonds

• break bonds in one mole of the substance

C.

break one mole of similar bonds

Bond energy : It may be defined as the energy released when two atoms get bonded or it is equal to the energy required to break the bond to form the neutral atoms .

e.g , H-H (g) $\to$ H + H , $∆$H = + 103 Kcal/mol

3.

The magnitude of orbital angular momentum of an electron of azimuthal quantum number 2 is :

• $\frac{2\mathrm{h}}{2\mathrm{\pi }}$

• $\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}$

• $\frac{\sqrt{2}\mathrm{h}}{2\mathrm{\pi }}$

• $\frac{6\mathrm{h}}{2\mathrm{\pi }}$

B.

$\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}$

Orbital angular momentum =  $\frac{\mathrm{h}}{2\mathrm{\pi }}$

where l = azimuthal quantum number

$\frac{\sqrt{6}\mathrm{h}}{2\mathrm{\pi }}$

4.

Heat of neutralisation of an acid with a base is 13. 7 kcal when :

• both acid and base are weak

• acid is weak and base is strong

• both acid and base are strong

• acid is strong and base is weak

C.

both acid and base are strong

Heat of neutralisation for strong acid and strong base is always close to 13.7 kcal .

+  $⇌$ H2O + AB

$∆$H = - 13.7 Kcal

The heat of neutralisation of strong acid and strong base is actually the heat of formation of 1 gmole of water from 1 g of H+ and 1 g of OH- .

5.

Hess's law is based on :

• law of conservation of mass

• law of conservation of energy

• enthalpy is a state function

• none of the above

B.

law of conservation of energy

Hess's law is based upon conservation of energy .

6.

Heisenberg's uncertainty principle can be explained as :

• $∆$$⩾$

• $\mathrm{∆}$x x $∆$$⩾$ $\frac{\mathrm{h}}{4\mathrm{\pi }}$

• $∆$x x $∆$$⩾$ $\frac{\mathrm{h}}{\mathrm{\pi }}$

• $∆$$⩾$ $\frac{\mathrm{\pi h}}{∆\mathrm{x}}$

B.

$\mathrm{∆}$x x $∆$$⩾$ $\frac{\mathrm{h}}{4\mathrm{\pi }}$

Heisenberg's uncertainty principle :  It is not possible to determine precisely both the position and the momentum (or velocity) of a small moving particle (e.g , electron , proton etc)

$∆$x . $∆$$⩾$ $\frac{\mathrm{h}}{4\mathrm{\pi }}$

where $∆$x , $∆$p are the uncertainties with regard to position , momentum respectively .

7.

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

C.

0.0087 g/L

Mg(OH)2 $⇌$ Mg2+ + 2OH-

Ksp = [Mg2+][OH-]2

1.4 X 10-11 = (s)(2s)2

1.4 X 10-11 = 4s3

s = 1.5 x 10-4 mol/L

($\because$Mol. wt. of Mg(OH)2 = 52)

= 58 X 1.5 x 10-4

= 0.0087 g/L

8.

The pH of a 0.02 M solution of HCl is :

• 2.2

• 2.0

• 0.3

• 1.7

D.

1.7

0.02 M HCl = 2 x 10-2 M HCl

[H+] = 2 x 10-2

pH = - log[H+]

= - log [2 x 10-2]

= - log 2 + 2 log 10

= - 0.3010 + 2

= 1.7

9.

The heat of formarion of CO2 is - 393 kJ mol-1 .The amount of heat evolved in the formation of 0.156 kg of CO2 .

• - 1393 kJ

• + 1165.5 kJ

• + 1275.9 kJ

• - 1165.5 kJ

A.

- 1393 kJ

$\because$ Heat evolved during the formation of 44 g of CO2 = - 93kJ

$\therefore$ Heat evolved in the formation of 0.156 kg of CO2

=

= - 1393 kJ

10.

The geometry of methane molecule is :

• tetrahedral

• pyramidal

• octahedral

• square planar

A.

tetrahedral

Methane (CH3) shows the sp3 hybridisation .So , the geometry of methane molecule is tetrahedral . 