Question

Assuming the density of glass is 1.5g/mL (8.0 g/mL calibration weights), what is the buoyancy corrected mass of the bottle + lid (c)? The mass of the bottle + lid is 20.7392

please answer asap

Answer 1

An object will weigh less in air than it will in a vacuum. Hence, we should expect that the mass of the bottle and lid will weigh more after the buoyancy correction.   In this case, there is a difference between the density of glass and the density of the calibration weights so we can make the correction for buoyancy.

This is given by the equation: W_v = W_a x [ 1 + (1/D_o - 1/D_w) x 0.0012] where W_v is the weight of the object in vacuum, W_a is the weight of the object in air, D_o is the density of the object and D_w is the density of the calibration weight. The number, 0.0012 refers to the density of air under normal lab conditions.

We have the following values given where W_a of bottle and lid = 20.7392 g, D_o for glass = 1.5 g/ml, D_w = 8.0 g/mL.

Plug these values into the equation and solve for W_v.

W_v = 20.7392 g x [ 1 + (1/1.5 - 1/8) x 0.0012]

W_v = 20.7526 g is the buoyancy corrected mass