The density of air is $0.001293{\text{g/cm}}^3$ at STP(atm). Which of the following statements is/are correct?

The correct option is B Molecular weight is $28.96$ amu
We know that: $\text{Molecular mass (Numerical value)}=2\times \text{Vapour density}$
At STP or under standard conditions of pressure and temperature, $22.4\text{L}$ of any gas is equal to 1 mole.
Therefore 1 mole = $22.4\text{L}$
The Volume of gas = $22400\text{mL}$
We know that $\text{mass}=\text{density}\times \text{volume}$
Therefore, rearranging the above equation
Mass = $0.001293\times 22400=28.96\text{g}$
Hence, molecular mass is 28.96
Also, we know that, $\text{Molecular mass (Numerical value)}=2\times \text{Vapour density}$
Rearranging the equation, we get,
$\text{Vapour density}=\frac{\text{Molecular mass (Numerical value)}}{2}$
$=\frac{28.96}{2}=14.48$