Chemistry - Mole Concept Question with Solution | TestHub

2.8 g of molecule, ${\mathrm{A}}_3$ density $=0.75\mathrm{g}/\mathrm{L}$ at NTP

∴ vol occupied by ${\mathrm{A}}_3=\frac{2.8}{0.75}\mathrm{L}$

∴ No of mole of ${\mathrm{A}}_3$ at NTP $=\frac{2.8}{0.75\times 22.4}$

∴ No of atoms at NTP $=\frac{2.8\times 3\times 6\times 10^{23}}{0.75\times 22.4}$

$=\frac{16.8\times 3\times 10^{23}}{\frac{3}{4}\times 22.4}$

$=3\times 10^{23}$

Similarly 11.2 g of ${\mathrm{B}}_2$, density $3\mathrm{g}/\mathrm{L}$ at NTP

∴ No of B atoms at NTP $=\frac{11.2\times 2\times 6\times 10^{23}}{3\times 22.4}$

$=2\times 10^{23}$

and 5.6 g of C, density is 1.5 g

No of C atoms at NTP $=\frac{5.6\times 6\times 10^{23}}{1.5\times 22.4}$

$=1\times 10^{23}$

∴ Total number of $\mathrm{A},\mathrm{B}$ and C atoms in the mixture

$=3\times 10^{23}+2\times 10^{23}+1\times 10^{23}$

$=6\times 10^{23}$

$\therefore =\mathrm{N}\times 10^{-23}=6\times 10^{23}\times 10^{-23}=6$