Question

The density of an ideal gas at $27{}^{o}C$ and pressure of 1 atm is $1.8\times {10}^{-3}g/c{m}^3,$ find the number of molecules in $90g$ of gas.

• A. $12\times {10}^{23}$
• B. $12\times {10}^{25}$
• C. $24\times {10}^{23}$
• D. $24\times {10}^{25}$

Answer 1

The correct option is A $12\times {10}^{23}$

$\frac{P}{\rho }=\frac{RT}{M}$
$M=\frac{RT\rho }{P}=\frac{8.3\times 300\times 1.8\times {10}^{-3}\times {10}^{-3}}{{10}^5\times {10}^8}=45g$
$n=\frac{90}{45}=2mole$
$N=2\times 6.02\times {10}^{23}=12\times {10}^{23}$