Question

An alloy of Iron$\left(50\%\right)$, nickel $\left(45\%\right)$ and manganese $\left(0.3\%\right)$ has a density of $8{\text{g/cm}}^3$. How many iron atoms are there on a block of alloy $10\text{cm}\times 20\text{cm}\times 15\text{cm}$?
$(\text{Avogardo's no}=N_A=6.023\times {10}^{23})$

• A. $1.29\times {10}^{26}$
• B. $3.62\times {10}^{26}$
• C. $1.56\times {10}^{26}$
• D. $2.44\times {10}^{26}$

Answer 1

The correct option is A $1.29\times {10}^{26}$

Volume of the block of alloy $=10\times 20\times 15{\text{cm}}^3$ $=3000{\text{cm}}^3$
Mass of the block $=3000\times 8\text{g}$ $=24000\text{g}$
Mass of the iron block $=\frac{50}{100}\times 24000$ $=12000\text{g}$
No of the iron atoms in block $=\frac{\text{Mass}}{\text{Atomic mass}}\times 6.023\times {10}^{23}$

$=\frac{12000}{56}\times 6.023\times {10}^{23}$ $=1.29\times {10}^{26}$