In a very good vacuum system in the laboratory, the vacuum atained was 10−13 atm. If the temperature the system was 300 K, the numbers of molecules present in a volume of 1cm3 is
A
2.4×106
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B
24
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C
2.4×109
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D
Zero
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Solution
The correct option is B2.4×106 From the Ideal gas Law we obtain, PV=nRT
⟹n= number of moles present in the given volume=PVRT
=(10−13×1.01×105)×(10−2)28.31×300
≈4.05×10−18
Thus number of molecules in the given volume=nNA=4.05×10−18×6.023×1023≈2.4×106