An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m-3, number of atoms present per unit cell is

Question

An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m-3, number of atoms present per unit cell is (A) 2 (B) 4 (C) 6 (D) 3

Tardigrade Admin · Accepted Answer

M (molar mass of the element) =2.7 × 10-2 kg mol-1 a (edge length) =405 pm=405 × 10-12m =4.05 × 10-10 m d (density) =2.7 × 103 kg m-3 NA (Avogadroâs number) =6.022 × 1023 mol-1 Using formula, Density d=(Z × M/a3× NA) or, Z=(d × a3× NA/M) or, Z=((2.7 × 103)(4.05× 10-10)3(6.022×1023)/2.7×10-2)=4 Number of atoms of the element present per unit cell = 4

Q. An element with molar mass $2.7 \times 1 0^{- 2} k g m o l^{- 1}$ forms a cubic unit cell with edge length $405 p m$ . If its density is $2.7 \times 1 0^{3} k g m^{- 3}$ , number of atoms present per unit cell is

$2$

$4$

$6$

$3$

Q. An element with molar mass 2.7×10−2kgmol−1 forms a cubic unit cell with edge length 405pm. If its density is 2.7×103kgm−3, number of atoms present per unit cell is

2

4

6

3

Q. An element with molar mass $2.7 \times 1 0^{- 2} k g m o l^{- 1}$ forms a cubic unit cell with edge length $405 p m$ . If its density is $2.7 \times 1 0^{3} k g m^{- 3}$ , number of atoms present per unit cell is

$2$

$4$

$6$

$3$