The molecular mass of $O_{2}$ and $S O_{2}$ are $32$ and $64$ respectively. If one litre of $O_{2}$ at $15^{\circ} C$ and $750 m m$ pressure contains $N$ molecules, the number of molecules in two litre of $S O_{2}$ under the same conditions of temperature and pressure will be:

2 N

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N

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\frac{N}{2}

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4 N

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Solution

The correct option is B $2 N$
From ideal gas equation, $P V = n R T .$
Oxygen, $V = 1 L, P = 750 m m H g$
Sulphur dioxide, $V = 2 L, P = 750 m m H g$
Moles of $O_{2} ⟹ \frac{P V}{R T} = n_{1} = \frac{1 \times 1}{R T}$
Number of molecules $= N = \frac{N a}{R T} ⟶ (i)$
Similarly for $S O_{2},$
Moles of $S O_{2} = \frac{P V}{R T} = n_{2} = \frac{1 \times 2}{R T}$
Let number of molecules $= M = \frac{2 N a}{R T} ⟶ (2)$
By dividing both $e q . 1$ and $e q . 2$ we get, $\frac{M}{N} = 2 M = 2 N$

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The molecular weight of $O_{2}$ and $S O_{2}$ are 32 and 64 respectively. If one litre of $O_{2}$ at $15^{\circ} C$ and 750 mm pressure contains 'N' molecules, the number of molecules in two litres of $S O_{2}$ under the same conditions of temperature and pressure will be