Chemistry

NEET Class 12

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1.

The cooling in refrigerator is due to :

the reaction of the refrigerator gas

the expansion of ice

the expansion of the gas in the refrigerator

the work of the compressor

C.

the expansion of the gas in the refrigerator

This is based on Joule-Thomson effect.

2.

A precipitate of AgCl is formed when equal volumes of the following are mixed :

[K_{sp} for AgCl = 10^{-10}]

10

^{-4}M AgNO_{3}and 10^{-7}M HCl10

^{-5}M AgNO_{3}and 10^{-6}M HCl10

^{-5}M AgNO_{3}and 10^{-4}M HCl10

^{-6}M AgNO_{3}and 10^{-6}M HCl

B.

10^{-5} M AgNO_{3} and 10^{-6} M HCl

For the precipitation of an electrolyte, it is necessary that the ionic product must exceed its solubility product. A precipitate of AgCl is formed when equal volumes of 10^{-5} M AgNO_{3 }and 10^{-5} M HCl are mixed, because ionic product will be 10^{-9} which is greater than K_{sp} (10^{-10}).

3.

Pick out the incorrect statement from the following :

sp hybrid orbitals are equivalent and are at an angle of 180° with each other

sp

^{2}hybrid orbitals are equivalent and bond angle between any two of them is 120°sp

^{3}d^{2 }hybrid orbitals are equivalent and are oriented towards corners of a regular octahedrondsp

^{2}hybrid orbitals are equivalent with a bond angle of 90° between the two atoms

D.

dsp^{2} hybrid orbitals are equivalent with a bond angle of 90° between the two atoms

Angle between any two dsp^{2} hybrid orbitals can be 90° or 180°

4.

An evacuated glass vessel weighs 50 g when empty, 144.0 g when filled with a liquid of density 0.47 g mL^{-}^{1}^{ }and 50.5 g when filled with an ideal gas at 760 mm Hg and 300 K. The molar mass of the ideal gas is :

[R = 0.0821 L atm K^{-1} mol^{-1}]

61.575

130.98

123.75

47.87

A.

61.575

Weight of empty glass vessel = 50 g

Weight of glass vessel filled with liquid = 144.0 g

$\therefore $Weight of liquid = 144- 50 = 94g

Density of liquid= 0.47 gmL^{-1}

$\therefore $Volume of liquid = $\frac{94}{0.47}$ = 200mL = 0.2L

Mass of glass vessel filled with gas= 50.5 g

$\therefore $ Mass of gas (m) = 50.5- 50 = 0.5 g

By using ideal gas equation ,

pV = $\frac{\mathrm{m}}{\mathrm{W}}$RT

where , p= 760 mm, Hg= 1atm ,

V = 0.2 L , T = 300K ,

and R = 0.0821 L atm K^{-1} mol^{-1}

$\therefore $Molar mass (M) =$\frac{\mathrm{mRT}}{\mathrm{pV}}$

= $\frac{0.5\times 0.0821\times 300}{1\times 0.2}$

= 61.575g mol^{-1}

5.

In an exothermic equilibrium ,

A + 3B $\rightleftharpoons $ AB_{3}

all the reactants and products are in gaseous state. The formation of AB_{3} is favoured at :

low temperature and low pressure

low temperature and high pressure

high temperature and high pressure

high temperature and low pressure

B.

low temperature and high pressure

The given reaction is exothermic ($\mathrm{\Delta}$H < 0) and accompanied by decrease in the number of moles. Hence, it will be favoured by low temperature and high pressure.

6.

A bivalent metal has an equivalent mass of 32. The molecular mass of the metal nitrate is :

182

168

192

188

D.

188

Atomic mass of the metal = 32 x 2 = 64

Formula of metal nitrate= M(NO_{3})_{2}

$\therefore $ Molecular mass = 64 + (14 + 3 x 16) x 2 = 188

7.

20 mL of 0.5 N HCl and 35 mL of 0.1 N NaOH are mixed. The resulting solution will :

be neutral

be basic

turn phenolphthalein solution pink

turn methyl orange red

C.

turn phenolphthalein solution pink

For 20 ml of 0.5 N HCl

N_{1}V_{1} = 0.5 x 20 = 10

For 35 ml of 0.1 NaOH

N_{2}V_{2} = 0.1x 35= 3.5

$\because $ N_{1}V_{1} > N_{2}V_{2}

$\therefore $So!ution will be acidic in nature.

Due to acidic nature of solution, the colour of phenolphthalein becomes pink.

8.

The line spectrum of He^{+} ion will resemble that of :

hydrogen atom

Li

^{+}ionhelium atom

lithium atom

A.

hydrogen atom

As hydrogen atom and He^{+} ion both have only one electron, hence the line spectrum of both will resemble.

9.

A body of mass x kg is moving with a velocity of 100 ms^{-1}. Its de-Broglie wavelength is 6.62 x 10^{-35} m. Hence, x is :

[h = 6.62 x 10^{-34}Js]

0.1 Kg

0.25 Kg

0.15 Kg

0.2 Kg

A.

0.1 Kg

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

$\therefore $ m = $\frac{6.62\times {10}^{-34}}{6.62\times {10}^{-35}\times 100}$

= 0.1 Kg

10.

0.5 mole of each of H_{2} , SO_{2} and CH_{4} are kept in a container. A hole was made in the container . After 3h , the order of partial pressures in the container will be :

pSO

_{2}> pH_{2}> pCH_{4}pSO

_{2}> PCH_{4}> pH_{2}pH

_{2}> pSO_{2}> pCH_{4}pH

_{2}> pCH_{4}> pSO_{2}

B.

pSO_{2} > PCH_{4} > pH_{2}

Rate of diffusion $\mathrm{\propto}$ $\frac{1}{\sqrt{\mathrm{molecular}\mathrm{mass}}}$

$\therefore $ Order of diffusion H_{2} > CH_{4} > SO_{2}

and amount left is in the order :

SO_{2} > CH_{4} > H_{2}

Hence, order of partial pressure is

pSO_{2} > pCH_{4} > pH_{2}

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