Chemistry

NEET Class 12

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

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

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 H_{2}O, 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 1s^{2}, 2s^{2}2p^{6}, 3s^{2}3p^{6}3d^{9} represents a

metal atom

non metal atom

non-metallic anion

metallic cation

D.

metallic cation

Electronic configuration of Copper (Cu) is 1s^{2}, 2p^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{1}, 3d^{10} and electronic configuration of Cu^{2+} is 1s^{2}, 2s^{2} 2p^{6}, 3s^{2} 3p^{6} 3d^{9}. 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 H

_{2}SO_{4}0.5 N H

_{2}SO_{4}1 N H

_{2}SO_{4}2 N H

_{2}SO_{4}

A.

4.0 N H_{2}SO_{4}

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 H_{2}SO_{4}, n = 2

So, 4.0 N H_{2}SO_{4} solution ≈ 2 M H_{2}SO_{4} 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 = $\frac{80}{20}\times 8$

$\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 PbSO_{4}, the volume of '10 volume' H_{2}O_{2} 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 + 4H_{2}O_{2} → PbSO_{4} + 4H_{2}O

$\because $ 1 mole of PbS requires 4 moles of H_{2}O_{2}.

$\therefore $ 0.01 moles of PbS requires 0.04 mole of H_{2}O_{2}.

Now,

Weight of 0.04 mole H_{2}O_{2} = 1.36 gm

And, 10 volume of H_{2}O_{2} means,

1 mL of such solution of H_{2}O_{2} on decomposition by heat produces 10 mL of oxygen at NTP.

H_{2}O_{2} decomposes itself as,

2H_{2}O_{2} → 2H_{2}O_{ }+ O_{2}

Thus, 1 mL of 10 volume of H_{2}O_{2} solution contains

= $\frac{68}{22400}$ × 10 gm of H_{2}O_{2} = 0.03035 gm H_{2}O_{2}

$\because $ 0.03035 gm of H_{2}O_{2} is present in 1 mL of 10 volume of H_{2}O_{2}

$\therefore $ 1.36 gm of H_{2}O_{2} present in $\frac{1}{0.03035}$ × 1.36 mL of 10 volume of H_{2}O_{2}

= 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 NH_{3}. 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^{-} + 2H_{2}O + NH_{3}

Now, from the equation, we get

8 moles of electron absorbed by 85 gm of NaNO_{3}.

$\therefore $ 1 mole of electron absorbed by $\frac{85}{8}\mathrm{g}$ of NaNO_{3} = 10.625 gm

Textbook Solutions | Additional Questions