$(a)\;2.68\times10^3\qquad(b)\;2.68\times10^4\qquad(c)\;2.68\times10^6\qquad(d)\;2.68\times10^{10}$

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Given

$P = 7.6\times10^{-10}mm$

$\;\;\;=\large\frac{7.6\times10^{-10}}{760}atm$

V = 1 litre

T = 273K

PV = nRT

$\large\frac{7.6\times10^{-10}}{760}\times1 = n\times0.0821\times273$

$ n= 4.46\times10^{-14} mole\;of\;O_2$

No. of molecules of $O_2 = 4.46\times10^{-14}\times6.023\times10^{23}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=2.68\times10^{10}$

Hence answer is (D)

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