what is the barometric pressure of CHBrCl2 (density= 1.971 g/cm3) 535.3 mm in height?

Answer 1

I don't understand the term barometric here. From the data you have given it seems you want the pressure at the bottom of a 535.3 mm column of a liquid with density = 1.971 g/cc. For such problems, pressure equals g * height * density. (Units check, using density in kg/m^3: N/kg * m * kg/m*3 = N/m^2 = Pa.)
P = 9.81 * 0.5353 * 1971 = 10350.3 Pa

Answer 2

To figure out the barometric pressure is if a column of CHBrCl2 is 535.3 mm in height, use the relationship
h1/h2 = d2/d1
where h is height in the column (diameter of the column doesn't matter) and d is the density of the material; 1 and 2 indicate the two different chemicals. The reason for this is that the same pressure will be exerted on the surface of the material; to exert an equal pressure back, less dense materials have to have a higher column than more dense materials.

To get barometric pressure, usually measured using mmHg (millimeters of mercury) or atm (atmospheres; 1 atm = 760 mmHg), convert your number into mmHg using the numbers given and the density of mercury (13.5 g/mL)

h(CHBrCl2)/h(Hg) = d(Hg)/d(CHBrCl2)

Solve for h(Hg)

h(Hg) = h(CHBrCl2)*d(CHBrCl2)/d(Hg)
=(535.3 mm)(1.971 g/mL)/(13.5 g/mL)
=78.2 mm Hg