Question

An investigational drug is about to enter clinical trials. It is insoluble in water, and in a powder form at room temperature. To determine the specific gravity of the powder, we will use a pycnometer that weighs 25.0 g when empty, and 106g when filled with water. When 16.0g of the powder is introduced to the bottle and the bottle is filled with water, the three together weigh 110g. What is the specifc gravity of the powder? (please explain)

Answer 1

m = 25g empty

m = 106 filled

X-X-X-X

m = 16 g powder

mt = 110g

Find s.g.

volume of water = mass/rho = (106-25)/1 = 81 ml (free volume) or volume used by water in the pycnometer

By definition, the total mass (we will read in the balance will be)

Mt = mw + mpowder + mpycnometer

mw = mass of water+ mass of poweder + mass of the pycnometer

Substitute values

110 = mw + 16 + 25

then solve for mw

mw = 69 grams of water (were poured)

assume density = 1g/ml

69 ml of only water

this means

Vw = 69 ml

The total volume - Volume of water (81-69) = 14

Then the density of the solid

density = mass/volume = 16/(81-69) = 1.33 g/ml

that is the density... but we need s.g. which is

s.g. = density of suybstance/ density of water = 1.33/1 = 1.33

s.g. = 1.33