AB crystallizes in a rock salt structures with A: B =1: 1. In shortest distances between A and B is Y1 / 3 nm. The formula mass of A B is 6.023 Y amu, where Y is arbitrary constant, find the density in kgm -3.

Question

AB crystallizes in a rock salt structures with A: B =1: 1. In shortest distances between A and B is Y1 / 3 nm. The formula mass of A B is 6.023 Y amu, where Y is arbitrary constant, find the density in kgm -3. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

a=2 × XY 1 / 3 nm ; M=(6.023 Y /1000) kgmol -1 ; Z =4 Density =(Z × M/a3 × N0) =(4 × 6.023 × 10-3 Y/(2 × Y1 / 3 × 10-9) × 6.023 × 1023) =5 kg m -3

Q. $A B$ crystallizes in a rock salt structures with $A : B = 1 : 1$ . In shortest distances between $A$ and $B$ is
$Y^{1/3} nm$ . The formula mass of $A B$ is $6.023 Y am u$ , where $Y$ is arbitrary constant, find the density in $k g m^{- 3}$ .

$a = 2 \times X Y^{1/3} nm$ ;
$M = \frac{6.023 Y}{1000} k g m o l^{- 1}; Z = 4$
Density $= \frac{Z \times M}{a ^{3} \times N _{0}}$
$= \frac{4 \times 6.023 \times 1 0 ^{- 3} Y}{( 2 \times Y ^{1/3} \times 1 0 ^{- 9} ) \times 6.023 \times 1 0 ^{23}}$
$= 5 k g m^{- 3}$

Q. AB crystallizes in a rock salt structures with A:B=1:1. In shortest distances between A and B is Y1/3nm. The formula mass of AB is 6.023Yamu, where Y is arbitrary constant, find the density in kgm−3.

a=2×XY1/3nm; M=10006.023Y​kgmol−1;Z=4 Density =a3×N0​Z×M​ =(2×Y1/3×10−9)×6.023×10234×6.023×10−3Y​ =5kgm−3

Q. $A B$ crystallizes in a rock salt structures with $A : B = 1 : 1$ . In shortest distances between $A$ and $B$ is
$Y^{1/3} nm$ . The formula mass of $A B$ is $6.023 Y am u$ , where $Y$ is arbitrary constant, find the density in $k g m^{- 3}$ .

$a = 2 \times X Y^{1/3} nm$ ;
$M = \frac{6.023 Y}{1000} k g m o l^{- 1}; Z = 4$
Density $= \frac{Z \times M}{a ^{3} \times N _{0}}$
$= \frac{4 \times 6.023 \times 1 0 ^{- 3} Y}{( 2 \times Y ^{1/3} \times 1 0 ^{- 9} ) \times 6.023 \times 1 0 ^{23}}$
$= 5 k g m^{- 3}$