1. D = (48.0 g) / (6.0 mL) = 8 g/mL  8.0 g/mL
QUESTION:
Illustrated below is a commonly used procedure for determining the density of an irregularly shaped, water-insoluble solid. The solid is weighed, then submerged in water. Suppose the solid weighed 48.0 g and the water level in the cylinder rose from 20.9 mL to 26.9 mL The density of the solid is... A g/mL, B g/mL C g/mL, D g/mL
D = m / V
V = 26.9 mL – 20.9 mL
= 6.0 mL
D = (48.0 g) / (6.0 mL) = 8 g/mL  8.0 g/mL
Illustrated in the figure is a commonly used procedure for determining the density of an irregularly shaped, water-insoluble solid. The solid is weighed, then submerged in water. Suppose the solid weighed 48.0 g and the water level in the cylinder rose from 20.9 mL to 26.9 mL The density of the solid is... A g/mL, B g/mL C g/mL, D g/mL
PAUSE
CLICK
Density is defined as the ratio of mass to volume.
Therefore, to calculate density, we simply divide the mass by the volume.
We are given that the mass is 48.0 grams.
HIGHLIGHT 48.0 g in problem
The volume of the object is equal to the volume of water displaced.
We calculate this by taking the difference between the final water level and initial water level in the cylinder.
CLICK The volume is 26.9 mL minus 20.9 mL,
CLICK or 6.0 mL
Therefore, the density is
CLICK the mass grams
HIGHLIGHT 48.0 g
divided by the volume mL HIGHLIGHT 8.0 mL which gives us 8 grams per milliliter.
Since 48.0 has 3 significant digits and 6.0 has two significant digits, the less precise term is 6.0
and we should express our answer to two significant digits.
Therefore, the correct answer is 8.0 grams per mL
The correct answer is C.
CLICK PAUSE END RECORDING

2. Video ID: 1A-8-3 © 2008, Project VALUE (Video Assessment Library for Undergraduate Education), Department of Physical Sciences Nicholls State University Author: Glenn V. Lo Narrator: Funded by Louisiana Board of Regents Contract No. LA-DL-SELECT-13-07/08