Question

An alloy of iron containing 54.7% iron and a Density of 8.17gcm-3 How many irons atoms are there in a alloy measuring 10cm × 15 cm × 20cm​

Answer 1

$\huge\mathfrak\red{solution}$

Volume of the block of alloy

=10cm×20cm×15cm=3000cm^3.

Density of the alloy is 8.17g/cm3.

Mass of the block of alloy .

=density×volume=8.17×3000=24510g.

The amount of iron present in the block of alloy =54.7/100×24510=13407g

The number of iron atoms present in the block =

13407/55.8×6.02×1023=1.442×1026.

Answer 2

$\huge\underline\purple{\sf Answer:-}$

★$\large{\boxed{\sf No.\:of\:atoms\:Fe=14.41×{10}^{25}}}$

$\huge\underline\purple{\sf Solution:-}$

★$\small{\boxed{\sf Volume\:Of\:Block = Lenght × breath × height}}$

$\large\implies{\sf 10×20×15}$

$\large\implies{\sf 3000{cm}^{3}}$

★Given :-

Volume (V ) = 3000${\sf {cm}^{3}}$

Density ${\sf \rho =8.17g{cm}^{-3}}$

★$\large{\boxed{\sf Mass\:of\:block=Volume×Density}}$

$\large\implies{\sf 3000×8.17}$

$\large\implies{\sf 24510g}$

$\large{\sf Mass\:of\:Fe\:block={\frac{24510×54.7}{100}}}$

$\large\implies{\sf 13406.7g}$

$\large{\boxed{\sf Mole(n)={\frac{Mass}{Molecular\:mass}}}}$

$\large{\sf No.\:of\:moles\:Fe={\frac{13406.6}{55}}}$

$\large\implies{\sf 239.41mol}$

$\large{\boxed{\sf No.\:of\:atom=n×N_{A}}}$

$\large{\sf N_{A}=6.022×{10}^{23}}$

$\small{\sf No.\:of\:atoms\:Fe=239.41×6.022×{10}^{23}}$

$\large\implies{\sf 14.41×{10}^{25}atoms}$

★$\large\red{\boxed{\sf No.\:of\:Fe\:atoms=14.41×{10}^{25}}}$