$\begin{array}{1 1}(a)\;6.022\times 10^{23}&(b)\;1.2044\times 10^{23}\\(c)\;2\;mole&(d)\;6.023\times 10^{24}\end{array}$

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44g $CO_2=N$molecules

$N=6.023\times 10^{23}$

$\therefore 4.4g CO_2=\large\frac{N}{44}$$\times 4.4$

$\Rightarrow \large\frac{N}{10}$

22.4 litres $H_2$ at STP=N molecules.

$\therefore$ 2.24 litres $H_2$ at STP$=\large\frac{N}{10}$molecules

Thus total molecule =$\large\frac{N}{10}+\frac{N}{10}$

$\qquad\qquad\quad\quad\quad=\large\frac{N}{5}$

(i.e) $\Rightarrow \large\frac{6.023\times 10^{23}}{5}$

$\Rightarrow 1.2044\times 10^{23}$

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