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**Hint:**The compressibility factor gives us an idea about the degree to which a real gas shows deviation from the ideal gas behaviour. It is the ratio of the observed molar volume of a gas to the calculated molar volume (using ideal gas equation) of the gas at the same pressure and temperature

**Complete step by step answer:**

-The compressibility factor for a gas is the ratio of the observed molar volume of a gas to the calculated molar volume (using ideal gas equation) of the gas at the same pressure and temperature.

- It can also be defined as the ratio of the product and pressure and volume of the gas to the product of the number of moles, the gas constant and the temperature of the gas.

-The equations are given below:

$Z=\dfrac{P V}{n R T}$

$\Rightarrow Z=\dfrac{P V_{\text {real}}}{n R T}$

(Where Z is the compressibility factor)

...(1)

Since according to the ideal gas equation: PV=nRT

Putting the above equation in equation (1),

Now, n is the number of moles which is given mass(m) divided by the molar mass (M).

$\mathrm{n}=\dfrac{\mathrm{m}}{\mathrm{M}}$

$\mathrm{Z}=\dfrac{\mathrm{PMV}}{\mathrm{mRT}}$

Now, we know that density () = $\mathrm{m} / \mathrm{V}$ ;

$\mathrm{Z}=\dfrac{\mathrm{PM}}{\rho \mathrm{RT}}$

Thus, now putting the values

P = 500 atm

M = 16 g/mol

= 0.246 g/ml

R = 8.314 L atm/mol/K

T = 200 K

Thus,

$\mathrm{Z}=\dfrac{16 \times 500}{0.246 \times 8.314 \times 200}$

$\mathrm{Z} \approx 20$

Thus Z = $2.0 \times 10^{1}$

**Clearly, the answer is B.**

**Note:**The compressibility factor for an ideal gas is 1 while for a real gas it could be less than 1 or greater than 1. If the compressibility factor is less than 1 then it implies that the attractive forces are predominant among the gas molecules. If the compressibility factor is more than 1 it implies that the repulsive forces are predominant among the gas molecules. Whether the compressibility factor will be greater than or less than 1 at a particular temperature and pressure will depend upon the nature of the gas.

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