Number of molecules in 1 litre of oxygen at NTP is:

\frac{6.02 \times 10^{23}}{32}

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\frac{6.02 \times 10^{23}}{22.4}

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32 \times 22.4

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\frac{32}{22.4}

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Solution

The correct option is A $\frac{6.02 \times 10^{23}}{22.4}$
At NTP, 1 mole of oxygen occupies a volume of 22.4 L.
Hence, 1 L of oxygen at NTP $= \frac{1}{22.4}$ mol.
The number of oxygen molecules $= \frac{1}{22.4} \times 6.02 \times 10^{23}$ $= \frac{6.02 \times 10^{23}}{22.4}$

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A mixture of CH₄, N₂ and O₂ is enclosed in a vessel of one litre capacity at 0⁰C. The ratio of partial pressures of gases is 1:4:2. Total pressure of the gaseous mixture is 2660 mm. The number of molecules of oxygen present in the vessel is

Q. Volume of a mixture of

6.02 \times 10^{23}

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Q. Statement 1: The number of gram molecules of oxygen in

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Statement 2: The value of Avogadro number is

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atoms and 1 mole =

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Number of molecules in 1 litre of oxygen at NTP is:

The correct option is A 6.02×102322.4At NTP, 1 mole of oxygen occupies a volume of 22.4 L.Hence, 1 L of oxygen at NTP =122.4 mol.The number of oxygen molecules =122.4×6.02×1023 =6.02×102322.4

The correct option is A $\frac{6.02 \times 10^{23}}{22.4}$
At NTP, 1 mole of oxygen occupies a volume of 22.4 L.
Hence, 1 L of oxygen at NTP $= \frac{1}{22.4}$ mol.
The number of oxygen molecules $= \frac{1}{22.4} \times 6.02 \times 10^{23}$ $= \frac{6.02 \times 10^{23}}{22.4}$