1

### JEE Main 2019 (Online) 10th January Evening Slot

A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms :
A
hcp lattice - A, ${1 \over 3}$ Tetrahedral voids-B
B
hcp lattice - B, ${1 \over 3}$ Tetrahedral voids - A
C
hcp lattice - A, ${2 \over 3}$ Tetrahedral voids - B
D
hcp lattice - B, ${2 \over 3}$ Tetrahedral voids - A

## Explanation

A2B3 has HCP lattice

If A form HCP,
then ${{{3^{th}}} \over 4}$ of THV must occupied by B to form A2B3

If B form HCP, then ${{{1^{th}}} \over 3}$ of THV must occupied by A to form A2B3
2

### JEE Main 2019 (Online) 11th January Morning Slot

An example of solid sol is :
A
Butter
B
Hair cream
C
Gem stones
D
Paint

## Explanation

An example of solid sol is gem stones.
3

### JEE Main 2019 (Online) 11th January Morning Slot

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
4

### JEE Main 2019 (Online) 11th January Evening Slot

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