Question

The number of molecules in $1{\text{cm}}^3$ of an ideal gas at $0^{\circ }$C and at a pressure of ${10}^{–5}\text{mm}$ of mercury is –
(Given that $760\text{mm}$ of mercury column measures $1\text{atm}$)

• A. $2.7\times {10}^{11}$
• B. $3.5\times {10}^{11}$
• C. $6.0\times {10}^{23}$
• D. $6\times {10}^{12}$

Answer 1

The correct option is B $3.5\times {10}^{11}$

Using ideal gas equation,
$PV=nRT\Rightarrow PV=\frac{N}{N_A}RT$
$\Rightarrow N=\frac{PV{N}_A}{RT}$
$N=\frac{[\frac{{10}^{-5}}{760}\times {10}^5]\times [1\times {10}^{-6}]\times 6\times {10}^{23}}{8.3\times 273}$
$N=3.5\times {10}^{11}$