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1

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
Given
Gas H2 CH4 CO2 SO2
Critical
Temperature / K
33 190 304 630



On the basis of data given above, predict which of the following gases shows least adsorption on a definite amount of charcoal ?
A
SO2
B
CH4
C
H2
D
CO2

Explanation

Adsorption $$ \propto $$ Tc

Therefore, H2 gas shows least adsorption on a definite amount of charcoal.
2

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
Among the colloids cheese (C), milk (M), and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is :
A
C : liquid in solid ; M : liquid in liquid ; S : solid in gas
B
C : liquid in solid ; M : liquid in solid ; S : solid in gas
C
C : solid in liquid ; M : liquid in liquid ; S : gas in solid
D
C : solid in liqid ; M : solid in liquid ; S : solid in gas

Explanation

Cheese (C) is liquid dispersed in solid.

Milk (M) is liquid dispersed in liquid.

Smoke (S) is solid dispersed in gas.
3

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is : (Edge length is represented by 'a')
A
0.134 a
B
0.067 a
C
0.047 a
D
0.027 a

Explanation

a = 2(R + r)

$$ \Rightarrow $$ $${a \over 2} = \left( {R + r} \right)$$ ......(1)

For bcc, $$\sqrt 3 a$$ = 4R

Using (i) and (ii)

$${a \over 2} = \left( {{{a\sqrt 3 } \over 4} + r} \right)$$

$$ \Rightarrow $$ r = $$a\left( {{{2 - \sqrt 3 } \over 4}} \right)$$

$$ \Rightarrow $$ r = 0.067 a
4

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)
A solid having density of 9$$ \times $$103 kg m–3 forms face centred cubic crystals of edge length $$200\sqrt 2 $$ pm. What is the molar mass of the solid?
[Avogadro constant $$ \cong $$ 6 $$ \times $$ 1023 mol–1 , $$\pi $$ $$ \cong $$ 3]
A
0.0305 kg mol–1
B
0.4320 kg mol–1
C
0.0216 kg mol–1
D
0.0432 kg mol–1

Explanation

$$\rho $$ = $${{Z \times M} \over {{a^3} \times {N_A}}}$$

$$ \Rightarrow $$ 9$$ \times $$103 = $${{4 \times M} \over {\left( {200\sqrt 2 \times {{10}^{ - 12}}} \right) \times 6 \times {{10}^{23}}}}$$

$$ \Rightarrow $$ M = 0.0305 kg mol–1

Questions Asked from Solid State & Surface Chemistry

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