# Number of ideal gas molecules per $cm^3$

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The lowest pressure that is the best vacuum that can be created in a laboratory at 27 degree celsius is $10^{-11}$millimetre of Mercury at this pressure the number of ideal gas molecules per $cm^3$ will be ?

As said ideal gas so I applied Ideal gas equation
$PV=nRT$
$[(10^{-11}×10^5)/(24.4×760)]×10^6$
this should give $n/V$
and multiplying it by avogadro number ahould give answer but it isn't ?

Symbols have there usual meanings

I am still not able to get the answer its given$3.22 × 10^5$ molecules/$cm^3$
I think it is the factor of $10^6$. This should be in the denominator, not the numerator.
okay  got it that was my mistake per $cm^3$ shoud be in denominator so $10^6$ would be in denominator ; okay got my error :)