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Standard VIII

Chemistry

Mixture

A mixture of ...

Question

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

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

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6.02 x 10²³

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22.4 x 10²²

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1000

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Solution

The correct option is A $\frac{6.02 \times 10^{23}}{22.4}$

Partial pressure of oxygen $= \frac{2}{1 + 4 + 2} \times 2660$
$= 760 m m$
Thus, $1 L$ oxygen gas is present at $0^{\circ} C$ and $760 m m$ pressure.
$∴ Number of oxygen molecules = \frac{6.023 \times 10^{23}}{22.4}$

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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

The correct option is A 6.02×102322.4 Partial pressure of oxygen =21+4+2×2660 =760 mm Thus, 1L oxygen gas is present at 0∘C and 760 mm pressure. ∴ Number of oxygen molecules =6.023×102322.4

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

The correct option is A $\frac{6.02 \times 10^{23}}{22.4}$

Partial pressure of oxygen $= \frac{2}{1 + 4 + 2} \times 2660$
$= 760 m m$
Thus, $1 L$ oxygen gas is present at $0^{\circ} C$ and $760 m m$ pressure.
$∴ Number of oxygen molecules = \frac{6.023 \times 10^{23}}{22.4}$