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
Comments
Leave a comment