Question

The specific gravity of the stainless steel spherical balls used in ball-bearings is $10.2$. How many iron atoms are present in each ball of diameter $1$ cm if the balls contain $84\%$ iron, by mass? (The atomic mass of iron is $56$.)

• A. $4.12\times {10}^{21}$
• B. $4.82\times {10}^{22}$
• C. $3.82\times {10}^{22}$
• D. None of the above

Answer 1

Answer: (B)

$4.82\times {10}^{22}$

Volume of sphere=$\frac{4}{3}\pi r^3$

Calculate Mass=?

Volume=?

Density=?

Since, specific gravity of iron=$\frac{densityofiron}{densityofwater}$

specific gravity of iron=$10.2=\frac{densityofiron}{0.999840}$

Density of iron $\simeq 10.2g/ml$

Volume$\frac{4}{3}\times 3.14\times {\left(0.5\right)}^3=0.52e{m}^3$

Mass=$density\times volume$

$10.2\times 0.52=5.34g$

Number of moles of iron=$mass/molarmass$

$\frac{5.34}{56}=0.095$

Given: The ball contains $84\%$ iron by mass

IRON atoms present in each ball

=$0.84\times 0.095\times 6.023\times {10}^{23}$

=$4.82\times {10}^{22}$