$\begin{array}{1,1}(a)\;3.29\times10^4 \;molecules\\(b)\;3.29\times10^6 \;molecules\\(c)\;3.29\times10^{10} \;molecules\\(d)\;3.29\times10^5 \;molecules\end{array}$

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For gaseous mixture at vaccum

$P = 10^{-10} mm =\large\frac{10^{-10}}{760} atm$

$V = 10^{-3} \;litre$

n = ?

R = 0.0821 litre atm $K^{-1}mol^{-1}$

T = 293 K

PV = nRT

$\therefore n = \large\frac{10^{-10}\times10^{-3}}{760\times0.0821\times293}$

$= 5.47 \times 10^{-18}$

$\therefore$ No. of molecules per mL $= 5.47\times10^{-18}\times6.023\times10^{23}$

$= 3.29 \times10^6$ molecules

Hence answer is (B)

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