Question

Which of the following contain maximum number of molecules at STP 100g of $CaC{O}_3$ 44.8 L of $O_3$, 2g of $H_2$ and 16g of $O_2$.

Answer 1

Step 1:

Maximum number of molecules

(i) 100 g of $CaC{O}_3$

Number of moles = $\frac{Mass}{molecularmass}$

$=\frac{100}{100}=1$

Number of molecules = $moles\times N_A$

where $N_A$ is Avogadro's number = $6.023\times {10}^{23}=1N_A$

Step 2:

(ii) 44.8 L of $O_3$

Number of moles = $\frac{Givenvolume}{22.4}$

$=\frac{44.8}{22.4}=2moles$

Number of molecules = $2\times N_A=2N_A$

Step 3:

(iii) 2 g of $H_2$

Number of moles = $\frac{2}{2}=1mole$

Number of molecules = $1N_A$

Step 4:

(iv) 16 g of $O_2$

Number of moles = $\frac{16}{32}=0.5moles$

Number of molecules = $0.5N_A$

Hence, $O_3$contains maximum number of molecules.