Question

An alloy of iron ($54.7\%$), nickel ($45\%$) and manganese ($0.3\%$) has a density of $8.17g/c{m}^3$. How many iron atoms are there in a block of alloy measuring $10cm\times 20cm\times 15cm$?

• A. $2.201\times {10}^{26}$
• B. $1.442\times {10}^{26}$
• C. $8.41\times {10}^{25}$
• D. $6.19\times {10}^{26}$

Answer 1

The correct option is B $1.442\times {10}^{26}$

Volume of the block of alloy $=10cm\times 20cm\times 15cm=3000c{m}^3$.

The density of the alloy is $8.17g/c{m}^3$.

Mass of the block of alloy $=density\times volume=8.17\times 3000=24510g$.

The amount of iron present in the block of alloy $=\frac{54.7}{100}\times 24510=13407g$

The number of iron atoms present in the block $=\frac{13407}{55.8}\times 6.023\times {10}^{23}=1.442\times {10}^{26}$.