{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Density of Liquids and Solids

Density of Liquids and Solids - Results and Discussion A...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
-1 Results and Discussion: A normal glass beaker was weighed on an electrical balance and found to weigh 21.601 g. The beaker was then filled with 10.0 mL of water and weighed to show the mass of 31.511 g. The mass of the 10.0 mL of water was found by subtracting the weight of the beaker (21.601 g) from the weight of the beaker and water (31.511 g) and found to be 9.91 g. The density of water was then calculated using Density=Mass/Volume. D=9.91g/10mL= .991 g/mL Percent error was then calculated by using the accepted value for the density of water (1.00 g/mL), and then equation: Percent Error=abs((accepted value-mean value)/(accepted value)) x 100 %=abs((1.00 g/mL -.991 g/mL)/1.00 g/mL)x 100= .6% This procedure was then repeated for the unknown liquid and sold. Unknown Liquid: clear, strong smelling, miscible with water Trial Beaker wt. Beaker+Liquid  Difference /10mL=Density 1 23.217 g 31.248 g 8.031 g .8031 g/mL 2 23.282 g 31.2969 g 8.014 g .8014 g/mL 3 23.313 g 31.339 g 8.026 g .8026 g/mL Based on these observations, Ethanol is assumed as unknown liquid. Trial Calculated Density Known Density Percent Error 1 .8031 g/mL
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}