CBSE Class 11-science Answered

Now 1 inch=2.54 cm

Radius of ball bearing is 01. inch =0.254 cm

Since the ball bearing is spherical in shape its volume is given by the formula 4/3 x 3.14 x r³

Since the radius of ball bearing is 0.254 cm ,its volume is 4/3 x 3.14 x (0.254)³ =0.0686 cm³

Further we are given the density of ball bearing as 7.75 g/cc

Therefore the mass for 0.0686 cm³ volume of ball bearing is 53.17 g

Now we are given that the stainless steel contains 85.6 % F by weight i.e. if the ball bearing weighs 100 g then it contains 85.6 g of Fe in it.

If 100 g of ball bearing contains 85.6 g Fe

Therefore 53.17 g of ball bearing would contain (85.6 x 53.17)/100=45.51 g Fe.

No. of moles of Fe= 45.51/56=0.812 mol

Now 1 mole of Fe contains 6.023 x 10²³ atoms of Fe

Therefore 0.812 mol of Fe contains 0.812  x 6.023 x 10²³ =4.98 x 10²³ atoms of Fe