DILIP
Last Activity: 8 Years ago
Given that the radius of the ball is 0.254 cm.
Volume of the ball is 4/3 π r3 .
Volume is 0.0686 cc.
mass of steel ball is = Density of ball x Volume of ball .
Mass = 7.75 x 0.0686 = 0.53165 g .
mass percent of Fe in ball is 85.6%
Mass of Fe in ball = ( 0.53165 x 85.6 ) / 100 = 0.455 g .
55.8 g of Fe = 1 mol of Fe .
0.455 g of Fe = 0.00815 moles of Fe.
1 mol of Fe contains ….. 6.023 x 1023 atoms.
0.00815 moles of Fe contains ….… ????
= ( 6.023 x 1023 ) x ( 0.00815 )
= 4.908 x 1021 iron atoms are present in the given stainless steel ball.
