Topics: Density, Concentration, Water Pages: 7 (1393 words) Published: January 27, 2015
Hannah Herbert
Density and Making Solutions
Objective and Background/ Theory of Experiment:
The objective of the experiment was to determine the density of a metal along with the density of distilled water. In an attempt to help the experimenters more thoroughly understand the relationship between concentration and density. Archimedes a Greek mathematician in the third century B.C. originally determined the relationship between the amount of matter that is within a particular space. He used this knowledge to determine if a crown made for the King was in fact made of pure gold, which as happens was not. He discovered such knowledge when he realized that the volume of an object could be determined by the displacement of the liquid. Density is a property that will be consistent for any substance. For pure substances it is the same concept as concentration. For solutions of material density increases as the concentration increases. Many substances can be found by density alone, and concentration can be identified if the density is known. In this experiment, the experimenters found the density of solid objects and then the concentrations of solutions based on their densities. It was determined that as density increases, concentration does as well. The density of a substance is something that stays constant, for solutions of material the density is expected to increase in direct translation to the concentration increasing. The theory surrounding this lab was to see in practice how density could be determined with displacement and then to determine whether the density does in fact increase directly with the concentration.

Summary of Key Data
Figure 1: From Part 1
Mass of Brass
Initial Volume
Finial Volume

Figure 2: From Part 2
Mass of the Empty 50-mL Flask
Weight of Flask Filled with 50-mL of Water
Weight Closest to 2.92g NaCl That Was Obtained
Mass of Filled Flask With the 1:1 Solution
Mass of Filled Flask With the 1:2 Solution
Mass of Filled Flask With the 1:10 Solution

Sample Calculations:
Part 1:
Volume of the Metal
Volume Metal= Volume final- Volume Initial
Volume final=18.45mL, Volume Initial=16.85mL
1.60mL= 18.45mL-16.85mL
Volume of Metal=1.60mL

Density of Metal
Density of Metal= Mass of Metal ÷ Volume of Metal
Mass of Metal = 13.22g Volume of Metal = 1.60mL
8.26g/mL = 13.22g ÷ 1.60mL
Density of Metal=8.26g/mL

Part 2:
Mass of Distilled Water
Mass of Water = Mass Filled – Mass Empty
Mass Filled= 87.0966g Mass Empty= 37.393g
49.704g=87.0966g- 37.393g
Mass of Distilled Water=49.704g

Density of Water
Density of Water= Mass of Water ÷ Volume of Water
Mass of Water= 49.7036g Volume of Water= 50mL
.99g/mL= 49.7036g ÷ 50mL
Density of Water=.99g/mL

Amount of NaCl for Stock Solution
Molarity (M) = moles solute ÷ Liters of Solution
Molarity= 10 M, moles of solute= x, Liters of Solution= 0.0500 x=(0.0500moles NaCl)(58.44g NaCl/moles NaCl)
x=2.92g NaCl

Actual Concentration Made
Weighed NaCl= 2.9412g
(2.9412g NaCl)(1mol NaCl/58.44g NaCl)=.503mol
Molarity= x÷ 0.0500L
1.006mol/L=.503mol÷ 0.0500L
Molarity= 1.006 mol/L

Mass of Stock Solution
Mass Solution= Mass filled2- Mass empty2
Mass filled 2=89.1207g, Mass empty 2=37.393g
51.728g=89.1207g- 37.393g
Mass Solution=51.728g

Density of Stock Solution
Density of Stock= Mass of Stock ÷ Volume of Stock
Mass Stock= 51.7277g, Volume of Stock= 50mL
Density of Stock Solution= 1.0g/mL

Calculation to find Volume of Stock Solution needed for the 1:2 Ratio Solution (M1)(V1)=(M2)(V2)
M1=1.006mol/L, V1=x, M2=.1006mol/L, V2=50mL
Volume needed=25mL

Mass of the 1:2 Solution
Mass of the Solution=Mass filled3- Mass empty
Mass filled3= 88.2571g, Mass empty= 37.393g
Mass of 1:2 Solution=...
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