Question.
Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of $\mathrm{Na}=23 \mathrm{u}, \mathrm{Fe}=56 \mathrm{u})$ ?
Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of $\mathrm{Na}=23 \mathrm{u}, \mathrm{Fe}=56 \mathrm{u})$ ?
Solution:
100 grams of sodium
No. of sodium atoms
$=\frac{\text { (Givenmass) }}{(\text { Gramatomic mass })} \times($ Avogadro's no. $)$
$=\frac{(100 g)}{(23 g)} \times\left(6.022 \times 10^{23}\right)=2.618 \times 10^{24}$ atoms
100 grams of iron
No. of iron atoms
$=\frac{\text { (Givenmass) }}{\text { (Gramatomic mass) }} \times($ Avogadro's no. $)$
$=\frac{(100 \mathrm{~g})}{(56 \mathrm{~g})} \times\left(6.022 \times 10^{23}\right)=1.075 \times 10^{24}$ atoms
100 g of sodium has more number of atoms
100 grams of sodium
No. of sodium atoms
$=\frac{\text { (Givenmass) }}{(\text { Gramatomic mass })} \times($ Avogadro's no. $)$
$=\frac{(100 g)}{(23 g)} \times\left(6.022 \times 10^{23}\right)=2.618 \times 10^{24}$ atoms
100 grams of iron
No. of iron atoms
$=\frac{\text { (Givenmass) }}{\text { (Gramatomic mass) }} \times($ Avogadro's no. $)$
$=\frac{(100 \mathrm{~g})}{(56 \mathrm{~g})} \times\left(6.022 \times 10^{23}\right)=1.075 \times 10^{24}$ atoms
100 g of sodium has more number of atoms
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