Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Ready2Learn on September 04, 2011, 10:14:46 PM
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Hello everyone, I just started taking a chemistry class and I'm struggling a bit (Admittedly I haven't taken a chemistry class in about 7 years). This is one of the post-lab questions I'm having some trouble with, I've started it, but not sure if I have done this right.
A graduated cylinder is filled to the 40.00mL mark with a mineral oil. The masses of the cylinder before and after the addition of the mineral oil are 124.966g and 159.446g, respectively. In a separate experiment, a metal ball bearing of mass 18.713 g is placed in the cylinder and the cylinder is again filled to the 40.00mL mark with mineral oil. The combined mass of the ball bearing and mineral oil is 50.952 g. Calculate the density and radius of the ball bearing (The volume of a sphere of radius r is (4/3)πr^3)
Experiment 1:
159.449g-124.966g=34.48g
34.48g=Mass of Mineral Oil
Density of Mineral Oil=M/V=34.48g/40mL=0.862g/cm^3
Experiment 2:
124.966+18.173=143.139=Mass of solid+cylinder
124.966+50.952=175.918=Mass of solid+ cylinder +Mineral Oil
175.918-143.139=32.779=Mass of added Mineral Oil.
Now I'm going to find the volume of the add mineral oil using the density from experiment one (I'm not sure if that's right)
V=32.779g/0.862 g/cm^3=38.02668213mL
Finding the volume of the solid, I subtract the added mineral oil volume from the total volume from experiment 1?
Volume of solid=40-38.02668213=1.97331787mL
Density of solid=18.713g/1.97331787mL=9.483013499g/cm^3
1.97331787=(4/3)(3.14)r^3
1.97331787*(3/4)(3.14)=.4713338861
.4713338861=r^3
r=0.7782328425158433
r=0.78
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you are perfectly on the right track
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you are perfectly on the right track
Thank you for your response... glad I wasn't completely off, lol.
So there's nothing here I need to fix, and those would be the correct answers for density and radius (when they're both rounded off)?
Thank you again