Question

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

• A. $1.422\times {10}^{26}$
• B. $1.422\times {10}^{23}$
• C. $6.23\times {10}^{23}$
• D. $1.656\times {10}^{26}$

Answer 1

The correct option is A $1.422\times {10}^{26}$

Volume of the block of alloy $=10\times 20\times 15{\text{cm}}^3=3000{\text{cm}}^3$
Mass of the block $=3000\times 8.17\text{g}=24510\text{g}$
Mass of iron in the block $=\frac{54.7}{100}\times 24510=13406.97\text{g}$
Number of atoms in the block $=\frac{\text{Mass}}{\text{Molar mass}}\times 6\times {10}^{23}$

$=\frac{13406.97}{56}\times 6\times {10}^{23}$
$=1.422\times {10}^{26}$