Answer
$h=9.22m$
Work Step by Step
We can determine the required height of column of ethylene glycol as follows:
First, we convert the given density into $Kg/m^3$
$d=\frac{1.12g}{cm^3}\times \frac{1Kg}{1000g}\times \frac{(100)^3cm^3}{(1)^3m^3}=1.12\times 10^3Kg/m^3$
Similarly, we convert Kilo Pascals into Pascals
$P=1atm\times \frac{101.325KPa}{1atm}\times \frac{1000Pa}{1KPa}$
$P=1.01325\times 10^5Pa$
We know that
$h=\frac{P}{dg}$
We plug in the known values to obtain:
$h=\frac{1.01325\times 10^5Pa}{(1.12\times 10^3Kg/m^3)(9.807m/s^2)}$
This simplifies to:
$h=9.22m$
