Answer
$20.1atm$
Work Step by Step
We can calculate the pressures (in atm) exerted by a column of isopropanol as follows:
First, we convert the given density into $Kg/m^3$
$d=\frac{0.785g}{cm^3}\times\frac{1Kg}{1000g}\times \frac{(100)^3cm^3}{(1)^3m^3}$
This simplifies to:
$d=785Kg/m^3$
Now $P=hdg$
We plug in the known values to obtain:
$P=264m\times 785Kg/m^3\times 9.807m/s^2$
This simplifies to:
$P=2.0324\times 10^6Pa$
This pressure in $atm$ can be converted as
$P=2.0324\times 10^6Pa\times\frac{1KPa}{1000Pa}\times \frac{1atm}{101.325KPa}$
This simplifies to:
$P=20.1atm$
