Question

1. Determine the volume that 1.33 X 10^24 iron atoms would occupy (d=7.87 g/cm^3)

Answer 1

QUESTION # 132011

Determine the volume that 1.33 X 10^24 iron atoms would occupy (d=7.87 g/cm^3)

ANSWER

1 mole of iron has mass of 55.845 g and 6.022×10^23atoms

6.022×10^23 atoms   have    55.845 g

1.33×10^24 atoms will have     ? g

(1.33×10^24 atoms × 55.845g) ÷ (6.022 ×10^23 atoms)

=123.338 g

Density of iron is 7.87 g/cm³ so,

7.87 g   has volume of   1 cm³

123.338 g  will have   ?  cm³

(123.338 g × 1 cm³) ÷ 7.87 g

=15.672 cm³