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Which has more number of atoms, $100\,g$ of sodium or $100\, g$ of iron (given, atomic mass of $Na = 23 \,u , \,Fe = 56 \,u $) ?
$100\, g$ of iron
Same in both
Data is not enough
$100\,g$ of sodium
Solution
Molar mass of sodium $=23 \,g$
$1$ mole atom $=6.022 \times 10^{23}$ atoms
$23 \,g$ sodium contains $=6.022 \times 10^{23}$ atoms
$1 \,g$ sodium contains $=6.022 \times 10^{23}$ atoms
$100\, g$ sodium contains $=6.022 \times 10^{23}$ atoms
$=2.618 \times 10^{24}$ atoms
By the above method or by formula we find number of atoms in $100\, g$
$Fe$ ;
Number of atoms of an element in a given mass
$=\frac {\rm {Given \,mass}}{\rm {Gram \,atomic\, mass \,100\,g}}$ $\times $ Avogadro's number
$=\frac{{100\,g}}{{56\,g}} \times \,6.022 \times {10^{23}}$
$ = \,1.075\, \times \,{10^{24}}$ atoms
Hence, $100\, g$ of sodium has more number of atoms as compared to $100\, g$ of iron.
