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# NEET Chemistry Solved Question Paper 2002

#### Multiple Choice Questions

1.

At 80°C, distilled water has H+ ions concentration equal to 1 x 10-6 mole/litre. The value of KW at this temperature will be :

• 1 X 10-6

• 1 X 10-12

• 1 X 10-9

• 1 X 10-15

B.

1 X 10-12

[H3O+] = 1 X 10-6 mol L-1

or [H+] = 1 X 10-6 mol L-1

As for distilled water [H+]= [OH]

[H+] = [OH-] = 1 X 10-6 mol L-1

Kw = [H+] [OH-] = 10-12 mol2 L2

2.

Spectrum of Li2+ is similar to that of :

• H

• Be

• He

• Ne

A.

H

Number of electrons in Li2+ is one similar to the hydrogen atom. One electron system shows similar kind of spectra. Many electron system spectra is complex due to inter-electronic repulsions.

3.

The pH of solution containing 0.10 M sodium acetate and 0.03 M acetic acid is :

(pKa for CH3COOH = 4.57)

• 4.09

• 6.09

• 5.09

• 7.09

C.

5.09

According to Henderson's equation,

pH = pKa + log $\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$

CH3COOH + NaOH $\to$ CH3COONa + H2O

Putting the values, we get :

pH = 4.57 + log $\frac{0.10}{0.03}$

$⇒$ pH = 5.09

4.

If P is pressure and $\mathrm{\rho }$ is density of a gas, then P and $\mathrm{\rho }$ are related as :

• $\propto$ $\mathrm{\rho }$

A.

$\propto$ $\mathrm{\rho }$

Ideal gas equation is given as :

PV = nRT

P = Pressure of gas, V = volume of gas

n = moles of gas, R = gas constant

T = temperature

PV = $\left(\frac{\mathrm{w}}{\mathrm{M}}\right)$RT  w = mass of gas , M = Molecular mass of gas

$⇒$ P =  $⇒$ P = $\mathrm{\rho }\frac{\mathrm{RT}}{\mathrm{M}}$

(As )

5.

The heat of neutralization of a strong base and a strong acid is 57 kJ. The heat released when 0.5 mole of HNO3 solution is added to 0.20 moles of NaOH solution, is :

• 11.4 kJ

• 34.7 kJ

• 23.5 kJ

• 58.8 kJ

A.

11.4 kJ

Heat of neutralization ($∆$H)= 57 kJ, mole of HNO3 = 0.5 mole and mole of NaOH = 0.2 mole. When HNO3 solution is added to NaOH solution, then 0.2 mole of HNO3 solution will combine with 0.2 mole of OH- ions of NaOH solution.

$\therefore$ Heat released= $∆$H x 0.2 = 57 x 0.2 = 11.4 kJ.

6.

Which of the following has maximum energy ?

• 3s                      3p                                     3d

 $⥮$ $⥮$ $↑$ $↑$

• 3s                      3p                                   3d

 $⥮$ $↑$ $↑$ $↑$ $↑$

• 3s                     3p                                     3d

 $⥮$ $↑$ $↑$ $↑$ $↑$ $↑$

• 3s                    3p                                 3d

 $⥮$ $↑$ $↑$ $↑$ $↑$

C.

3s                     3p                                     3d

 $⥮$ $↑$ $↑$ $↑$ $↑$ $↑$

The order of increasing energy of the subatomic orbitals is s < p < d < f. The energy in excited state is more than that in the ground state. Since option (c) is in an excited state, therefore it has maximum energy.

7.

Which of the following molecule has highest bond energy ?

• F-F

• N-N

• C-C

• O-O

C.

C-C

The bond energies of F-F , C - C , N - N and O-O bonds are 33, 80, 39 and 34.2 kJ/mol respectively. Therefore molecule of C-C has the highest bond energy.

8.

The weight of one molecule of a compound C60H120 is :

• 1.2 x 10-20 g

• 5.025 x 1023 g

• 1.4 x 10-21 g

• 6.023 x 1023 g

C.

1.4 x 10-21 g

Molecular weight of C60H122 = 60 x 12 + 1 x 122

= 842 g mol-1

$\therefore$ Weight of I mole of C60H122 molecules = 842 g

Weight of 6.023 x 1023 molecules = 842 g

Weight of 1 molecule =

9.

Quantum numbers of an atom can be defined on the basis of :

• Hund's rule

• Pauli's exclusion principle

• Aufbau's principle

• Heisenberg's uncertainty principle

B.

Pauli's exclusion principle

Each electron in an atom is designated by a set of four quantum numbers. According to Pauli's exclusion principle, no two electron in an atom have same values ofall the four quantum numbers. Therefore consequently, an orbital accommodates two electrons with opposite spins, these two electrons have the me value of quantum number 'n' , 'l' and 'm' but value of 's' will be different.

10.

The solubility of CuBr is 2 x 10-4 mol/L at 25°C. The KSP value for CuBr is :

• 4 x 10-8 mol2 L-2

• 4 x 10-4 mol2 L-2

• 4 x 10-11 mol2 L-2

• 4 x 10-15 mol2 L-2

A.

4 x 10-8 mol2 L-2

CuBr ➔ Cu+ + Br

Solubility of Cu Br is 2 x 104 mol L-1

Therefore, solubility of Cu+ = 2 x 10-4 mol L-1

Solubility product = 2 x 10-1 x 2 x 10-4

4 x 10-8 mol2 L-2