The molecular weight of $O_{2}$ and ${S O}_{2}$ are 32 and 64 respectively. If one litre of $O_{2}$ at $15$ and 759mm pressure contains N molecules, the number of molecules in two litre of ${S O}_{2}$ under the same conditions of temperature
(1) N/2 (2) N (3) 2N (4) 4N

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Solution

The correct is option is 3
$P V = n R T$

or , $\frac{P}{R T} = \frac{n}{v}$

since P, R and T is the same for both the gases, we can write

$(\frac{n}{v}) O_{2} = (\frac{n}{v} S O_{2})$
or, $\frac{n 0_{2}}{1} = \frac{n_{S} 0_{2}}{2}$
or $n_{o_{2}}$ = $\frac{n_{s o_{2}}}{2}$ or $n_{s o_{2}}$ = $2 n o_{2}$

for $O_{2}$

$N u m b e r o f m o l e s = \frac{n o . o f m o l e s}{6.23 \times 10^{2} 3} = \frac{N}{6.023 \times 10^{2} 3}$
Therefore,

Number of moles of $S o_{2}$ = $\frac{2 N}{6.23 \times 10^{2} 3}$
Now, no of molecules for $S o_{2}$

$n o o f m o l e c u l e s = n o o f m o l e s \times 6.23 \times 10^{2} 3$

$ number of molecules=\frac{2N}{6.023\times 10^23}\times 6.023\times
10^23$
$N o o f m o l e c u l e s o f S o_{2} = 2 N$

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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