Question

how many Iron atoms are present in stainless steel ball bearing having radius of 0.25 4 cm the stainless steel contains 80 5.6% ft by weight and density of 7.75 gram per centimetre cube​

Answer 1

Hey Dear,

◆ Answer -

Fe atoms = 4.897×10^21 atoms

● Explaination -

# Given -

r = 0.254 cm

d = 7.75 g/cm^3

# Solution -

Volume of the steel ball is calculated as -

V = 4/3 π × r^3

V = 4/3 × 3.142 × 0.254^3

V = 0.06865 cm^3

Mass of the steel ball will be thus -

W = V × d

W = 0.06865 × 7.75

W = 0.532 g

Mass of iron in the steel ball -

W1 = w/w % × W

W1 = 85.6/100 × 0.532

W1 = 0.4554 g

No of iron atoms present in that mass is given by -

N = NA × W1 / M1

N = 6.022×10^23 × 0.4554 / 56

N = 4.897×10^21 atoms

Therefore, given steel ball consists of 4.897×10^21 atoms.

Thanks for asking...