$4.4 \, \text{g}$ of $CO_2$ and 2.24 litre of $H_2$ at STP are mixed in a container. The total number of molecules present in the container will be:
A$6.022 \times 10^{23}$
B$1.2044 \times 10^{23}$
C$6.022 \times 10^{22}$
D$6.023 \times 10^{24}$
Answer & Solution
Correct answer: B. $1.2044 \times 10^{23}$
Moles of $CO_2 = \dfrac{4.4}{44}=0.1$. At STP, 2.24 L of $H_2$ corresponds to $\dfrac{2.24}{22.4}=0.1$ mol. Total moles of molecules $=0.2$, so total number of molecules $=0.2\times 6.022\times10^{23}=1.2044\times10^{23}$.
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