$0.1$ mol of $C H_{4} + 3.01 \times 10^{23}$ of molecules $C H_{4} - 9.6$ g of $C H_{4} = x$ mol of $H$ atoms. Here, the value of $x$ is:

0

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Solution

The correct option is A $0$
$3.01 \times 10^{23}$ molecules of $C H_{4} = \frac{3.01 \times 10^{23} C H_{4}}{6.02 \times 10^{23} C H_{4}} = 0.5$ mol $C H_{4}$
$9.6 g C H_{4} = \frac{9.6 g C H_{4}}{16 g C H_{4}} = 0.6$ mol $C H_{4}$
Hence, $0.1$ mol $C H_{4} + 3.01 \times 10^{23}$ molecules $C H_{4} - 9.6$ g $C H_{4}$
$= 0.1$ mol $C H_{4} + 0.5$ mol $C H_{4} - 0.6$ mol $C H_{4}$
$= 0.6$ mol $C H_{4} - 0.6$ mol $C H_{4} = 0$ mol $C H_{4}$
$= 0$ mol $H$ atoms
Hence, the value of $x$ is $0$ .

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0.1 mol of CH4+3.01×1023 of molecules CH4−9.6 g of CH4=x mol of H atoms. Here, the value of x is:

$0.1$ mol of $C H_{4} + 3.01 \times 10^{23}$ of molecules $C H_{4} - 9.6$ g of $C H_{4} = x$ mol of $H$ atoms. Here, the value of $x$ is: