## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# NEET Chemistry Solved Question Paper 2007

#### Multiple Choice Questions

1.

Heisenberg's uncertainty principle can beexplained 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 notpossible to determine precisely both theposition and the momentum (or velocity) of asmall moving particle (e.g , electron , proton etc)

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

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

2.

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 .

3.

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

$\mathrm{\nu }$=$\frac{{\mathrm{mc}}^{2}}{\mathrm{h}}$=$\frac{1.1×{10}^{-23}×3×{10}^{8}}{6.6×{10}^{-34}}$

= 5 x 1018 Hz

4.

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 energyreleased when two atoms get bonded or it isequal to the energy required to break the bondto form the neutral atoms .

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

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.

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 strongbase is always close to 13.7 kcal .

$\underset{\mathrm{strong}\mathrm{acid}}{\mathrm{HA}}$+$\underset{\mathrm{strong}\mathrm{base}}{\mathrm{BOH}}$$⇌$H2O + AB

$∆$H = - 13.7 Kcal

The heat of neutralisation of strong acid andstrong base is actually the heat of formation of1 gmole of water from 1 g of H+ and 1 g of OH- .

7.

The heat of formarion of CO2 is - 393 kJ mol-1 .The amount of heat evolved in the formation of0.156 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 ofCO2= - 93kJ

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

=$\frac{-393×156}{44}$

= - 1393 kJ

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 solubility product of Mg(OH)2 at 25°C is1.4 x 10-11 .What is the solubility of Mg(OH)2in 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

# 10.The magnitude of orbital angular momentum ofan 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 =$\sqrt{\mathrm{l}\left(\mathrm{l}+1\right)}$$\frac{\mathrm{h}}{2\mathrm{\pi }}$

where l = azimuthal quantum number

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