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Chapter 1: Problem 80

A pycnometer (see Exercise 78 ) weighs 25.60 g empty and \(35.55 \mathrm{g}\) when filled with water at \(20^{\circ} \mathrm{C}\) The density of water at \(20^{\circ} \mathrm{C}\) is \(0.9982 \mathrm{g} / \mathrm{mL}\). When \(10.20 \mathrm{g}\) lead is placed in the pycnometer and the pycnometer is again filled with water at \(20^{\circ} \mathrm{C}\), the total mass is \(44.83 \mathrm{g}\). What is the density of the lead in grams per cubic centimeter?

Short Answer

Expert verified
The density of the lead is \( 1.10 g/mL \) or \( 1.10 g/cm^{3} \)

Step by step solution

01

Determine volume of the pycnometer

The volume of the pycnometer can be established by utilising the recorded mass of the water and the known density of water. Given the mass of the water-filled pycnometer less the mass of the empty pycnometer equals to the mass of the filled water, which is \(35.55 g - 25.60 g = 9.95 g\). Now, use the mass and density of the water to find the volume. The volume will be \( \frac{mass_{water}}{density_{water}} = \frac{9.95 g}{0.9982 g/mL} = 9.97 mL \)
02

Determine volume of the lead

Subtract the mass of the pycnometer containing water and lead by the mass of the empty pycnometer and the mass of water. That gives the mass of the lead \(44.83 g - 25.60 g - 9.95 g = 9.28 g \). The difference between the total volume of water and pycnometer and the volume of the water will give the volume of lead. Therefore the volume of the lead is \(44.83 g - 35.55 g = 9.28 g \)
03

Calculate density of the lead

The density can be calculated by taking the mass of the lead divided by its volume. Therefore the density of the lead will be \( \frac{mass_{lead}}{volume_{lead}} = \frac{10.20 g}{9.28 mL} = 1.10 g/mL \)

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Most popular questions from this chapter

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