91Ó°ÊÓ

1 Pascal is the pressure exerted on a surface when a force of 1 Newton acts on a surface area of 1m².

$\text{Pressure} = \frac{\text{Force}}{\text{Area}}$

$1 \text{Pa} = \frac{\text{1} \text{N}}{\text{1} {\text{m}}^{\text{2}}}$

According to the given data, F = mg

Where,

F = force

m = mass

g = acceleration due to gravity

1 atm = $1.01325 × {10}^{5 }\text{Pa}$

Now, the mass of the atmosphere is,

$\begin{array}{rcl}1.01325 × {10}^5 \text{Pa} & =&  \frac{\text{mg}}{\text{area}}\\ 1.01325 × {10}^5 \text{Pa} & =&  \frac{\text{m}×\text{9.8} {\text{m/s}}^{\text{2}}}{\text{1} {\text{m}}^{\text{2}}}\\ \text{m} & =&  1.09 ×{10}^4\text{kg}\\ & & \end{array}$

Thus, the mass of the atmosphere is$1.09 × {10}^4 \text{kg}$.

The relation between density and pressure is,

$P=pgh$

Where,

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{ÒÏ} = \text{density}  \\ \text{g} =\text{acceleration} \text{due} \text{to} \text{gravity} \\ \text{h} = \text{height} \end{gathered}$$

Now, the height of the column is,

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{h}=\frac{\mathrm{P}}{\mathrm{pg}}=\frac{1.01325×{10}^5 \text{Pa}}{22.6 \text{²µ/³¾±ô×}9.81  {\text{m/s}}^{\text{2}}} \\ \text{h} = 0.457 \text{m} \end{gathered}$$

Thus, the height of the column is$0.457 \text{m}$