Question

) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at 25 °C, and the density of water at 25 °C is 0.9970 g>mL)
At the left, grains of the mineral calcite float on the surface of the liquid bromoform (d = 2.890 g/mL) At the right, the grains sink to the bottom of liquid Chloroform (d = 1.444 g/mL). By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined

Answer 1

Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

$m_{calcite}=15.4448g-12.4631g=2.9817g$

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

$V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL$

Thus, the density of the calcite sample will be:

$\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL$

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

Learn more:

• brainly.com/question/12001845
• brainly.com/question/11242138

Regards!