A mixture having 2 g of hydrogen and 32 g of oxygen occupies how much volume at NTP?

44.8 L

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22.4 L

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11.2 L

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67.2 L

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Solution

The correct option is C 44.8 L
Hydrogen exist as $H_{2}$ and oxygen as $O_{2}$
1 mole of all gases occupy 22.4 L volume at NTP
Moles in 2 g hydrogen = $\frac{m a s s}{m o l a r m a s s}$
$= \frac{2}{2} = 1$ mol=22.4 L
Moles in 32 g oxygen = $\frac{m a s s}{m o l a r m a s s}$
$= \frac{32}{32} = 1$ mol=22.4 L
Total volume occupied $= 22.4 + 22.4 = 44.8$ L

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