$\begin{array}{1 1}(a)\;22.4\times 10^3mL\;of\;CO_2\;at\;NTP\\ (b)\;22g\;of\;CO_2\;gas \\(c)\;11.2L\;of\;CO_2\;gas\;at\;NTP\\(d)\;0.1\;mole\;of\;CO_2\;gas\end{array} $

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22400ml of $CO_2$ gas =$6.023\times 10^{23}$molecules at NTP

$\therefore$ 44g of $CO_2$ gas =$6.023\times 10^{23}$ molecules

22g of $CO_2$ gas =$\large\frac{6.023\times 10^{23}}{44}$$\times 22$

$\Rightarrow 3.0115\times 10^{23}$ molecules

22.4L of $CO_2$ gas =$6.023\times 10^{23}$ molecules at NTP.

$\therefore$ 11.2L of $CO_2$ gas =$\large\frac{6.023\times 10^{23}}{22.4}$$\times 11.2$

$\Rightarrow 3.0115\times 10^{23}$ molecules

1 mole of $CO_2$ gas at NTP=$6.023\times 10^{23}$ molecules

0.1 mole of $CO_2$ gas at NTP=$\large\frac{6.023\times 10^{23}}{1}$$\times 0.1$

$\Rightarrow 6.023\times 10^{22}$ molecules

Hence (d) is the correct option.

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