the density of copper is listed as 8.94 g cm3

The density of copper is listed as 8.94 g/cm3. Two students each make three density determinations of samples of the substance. Student A’s results are 7.3 g/mL, 9.4 g/mL, and 8.3 g/mL. Student B’s results are 8.4 g/cm3, 8.8 g/cm3, and 8.0 g/cm3. How accurate and precise are the two sets of results?

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Lulama KlaasLv10

4 Nov 2020

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This question has multiple parts. Work all the parts to get the most points.

You want to determine the density of a compound but have only a tiny crystal, and it would be difficult to measure mass and volume accurately. There is another way to determine density, however, called the flotation method. If you placed the crystal in a liquid whose density is precisely that of the substance, it would be suspended in the liquid, neither sinking to the bottom of the beaker nor floating to the surface. However, for such an experiment, you would need to have a liquid with the precise density of the crystal. You can accomplish this by mixing two liquids of different densities to create a liquid having the desired density. Consider the following: you mix 8.90 mL of CHCl₃ (d = 1.492 g/mL) and 6.70 mL of CHBr₃ (d = 2.890 g/mL), giving 15.6 mL of solution. What is the density of this mixture?

g/mL

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Mass CHCl₃ = 8.90 mL Ã 1.492 g/mL = 13.3 g

Mass CHBr₃ = 6.70 mL Ã 2.890 g/mL = 19.4 g

Mass of solution = 13.3 g + 19.4 g = 32.6 g

Volume of solution = 8.90 mL + 6.70 mL = 15.6 mL

(It is important that you postpone rounding until all the calculations are completed. At least one extra digit beyond the significant digits should be carried through all of the computations in order to avoid a rounding error. Round only the final answer to the correct number of significant digits.)

Suppose now that you wanted to determine the density of a small yellow crystal to confirm that it is silicon. From the literature, you know that silicon has a density of 2.33 g/cm³. How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of CHCl₃ (d = 1.492 g/mL) and CHBr₃ (d = 2.890 g/mL)? (Note: 1 mL = 1 cm³.)

V(CHCl₃) = mL

V(CHBr₃) = mL