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# NEET Chemistry Solved Question Paper 2008

#### Multiple Choice Questions

1.

Unusually high boiling point of water is result of

• intermolecular hydrogen bonding

• intramolecular hydrogen bonding

• both intra and inter molecular hydrogen bonding

• high specific heat

A.

intermolecular hydrogen bonding

Due to presence of intermolecular hydrogen bonding in H2O, its molecules are associated with each other, which results unusual high boiling point of water.

2.

In a given shell the order of screening effect is

• s > p > d > f

• s > p > f > d

• f > d > p > s

• s < p < d < f

A.

s > p > d > f

The order of screening effect is s > p > d > f. Electrons experience the greatest effective nuclear charge when in s-orbital, then p-orbital and so on. Ionisation energy increases with an increase in penetration power.

3.

The electronic configuration 1s2, 2s22p6, 3s23p63d9 represents a

• metal atom

• non metal atom

• non-metallic anion

• metallic cation

D.

metallic cation

Electronic configuration of Copper (Cu) is 1s2, 2p2, 2p6, 3s2, 3p6, 4s1, 3d10 and electronic configuration of Cu2+ is 1s2, 2s2 2p6, 3s2 3p6 3d9. Hence, the given configuration represents metallic cation.

4.

A catalyst is a substance which

• increases the equilibrium constant of the reaction

• increases equilibrium concentration of products

• does not alter the reaction mechanism

• changes the activation energy of the reaction

D.

changes the activation energy of the reaction

A catalyst changes the activation energy of the reaction. As a result, the reaction follows an alternate path and the rate of reaction wlll change.

5.

2 N HCl solution will have same molar concentration as a

• 4.0 N H2SO4

• 0.5 N H2SO4

• 1 N H2SO4

• 2 N H2SO4

A.

4.0 N H2SO4

Normality = n × molarity (M)

where, n = the number of moles of H+ per mole of the compound that a solute is capable of releasing [acid] on reacting with base.

In case of HCl, n = 1

So, 2N HCl solution ≈ 2 M HCl solution

And in case of H2SO4, n = 2

So, 4.0 N H2SO4 solution ≈ 2 M H2SO4 solution

6.

An unknown element forms an oxide.What will be the equivalent weight of the element if the oxygen content is 20% by weight?

• 16

• 32

• 8

• 64

B.

32

According to the question, oxygen contents in element oxide is 20% by weight.

Hence, element contents in element oxide is 80% by weight.

Then, equivalent weight of unknown element = $\frac{80}{20}×8$

$\therefore$ Equivalent weight of unknown element = 32

7.

Which of the following expressions gives the de-Broglie relationship?

• $\mathrm{p}=\frac{\mathrm{h}}{\mathrm{mv}}$

• $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$

• $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mp}}$

• $\mathrm{\lambda m}=\frac{\mathrm{v}}{\mathrm{p}}$

B.

$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$

The de-Broglie relation is

$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$

where, λ = de-Broglie wavelength; h = Planck's constant; m = mass of particle; v = velocity of particle

8.

In transforming 0.01 mole of PbS to PbSO4, the volume of '10 volume' H2O2 required will be

• 11.2 mL

• 22.4 mL

• 33.6 mL

• 44.8 mL

D.

44.8 mL

From the equation given below,

PbS + 4H2O2 → PbSO4 + 4H2O

$\because$ 1 mole of PbS requires 4 moles of H2O2.

$\therefore$ 0.01 moles of PbS requires 0.04 mole of H2O2.

Now,

Weight of 0.04 mole H2O2 = 1.36 gm

And, 10 volume of H2O2 means,

1 mL of such solution of H2O2 on decomposition by heat produces 10 mL of oxygen at NTP.

H2O2 decomposes itself as,

2H2O2 → 2H2O + O2

Thus, 1 mL of 10 volume of H2O2 solution contains

$\frac{68}{22400}$ × 10 gm of H2O2 = 0.03035 gm H2O2

$\because$ 0.03035 gm of H2O2 is present in 1 mL of 10 volume of H2O2

$\therefore$ 1.36 gm of H2O2 present in $\frac{1}{0.03035}$ × 1.36 mL of 10 volume of H2O2

= 44.81 mL

9.

Addition of sodium acetate to 0.1 M acetic acid will cause

• increase in pH

• decrease in pH

• no change in pH

• change in pH that cannot be predicted

A.

increase in pH

Addition of sodium acetate in acetic acid solution, due to common ion NH${}_{4}^{+}$, The supressed, ionisation of acetic acid so concentration of [H+] decreases. Hence, pH of solution increases.

10.

Sodium nitrate on reduction with Zn in presence of NaOH solution produces NH3. Mass of sodium nitrate absorbing 1 mole of electron will be

• 7.750

• 10.625

• 8.000

• 9.875

B.

10.625

In the equation,

Zn + 2OH- → ZnO${}_{2}^{2-}$ + 2H+ + 2e-

NO${}_{3}^{-}$ + 8H+ + 8e- → OH- + 2H2O + NH3

Now, from the equation, we get

8 moles of electron absorbed by 85 gm of NaNO3.

$\therefore$ 1 mole of electron absorbed by $\frac{85}{8}\mathrm{g}$ of NaNO3 = 10.625 gm