Specific Gravity question

[jammin06]

hey guys, can someone explain specific gravity to me? All I understand about the concept is that it's a comparison of the density of some object compared to the density of water. However, I came across the following in a practice test and I am unable to solve the problem. Any help would be greatly appreciated. thanks

question:
A 12-g object dropped into a container of ethyl alcohol (S.G = 0.8) has an apparent mass of 8 g. What is the approximate density of the object?

A) 1.0 g/cm?
B) 1.6 g/cm?
C) 2.4 g/cm?
D) 5.0 g/cm?

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[willthatsall]

I think it should be C, 2.4 g/cm^3.

Apparent weight is basically the weight of the object minus the buoyant force of the ethanol. Since you know the ethanol has a density of .8 g/cm^3, you can figure out the volume of the displaced ethanol by using the difference between the apparent mass and the actual mass (4 g). This is because the mass of the displaced ethanol is equal to the difference between the apparent mass and the actual mass. So since the difference is (12-8=4g), that means 4g (5cm^3, using the density) of ethanol was displaced. That means the volume of the object was also 5cm^3. Just divide mass/volume and there you are (12g/5cm^3 = 2.4 g/cm^3). I think that is right, hope it helps.

Edited to add: I'm not sure if it was given in the problem, but all you really needed to know was the density of water, which would have told you the density of ethanol and then it's pretty straightforward.

 

[N-toxicologist]

weight of object/buoyant force = dens.obj/ dens. fluid

wt. obj = .12N
buoyant force = .04N (diff in apparent vs. actual wt.)
so this ratio = 3

dens. obj/ dens. fluid (which we know is .8) = 3
so dens obj. = 2.4 (I think)

What's the real answer?

If I'm wrong, I hope someone will tell me why. I probably need to know this...

 

[ken37]

2.4 is the answer

I use your ratio logic UNTLabrat, because it seems faster and more intuitive.

 

[Shrike]

All your math is right, but I think we can make it even easier:

If the object is 12g, and its apparent mass is 8g, then the difference, 4g, is the fluid displaced. It's 8/10ths the density of water, so the same volume would be 5g of water. 12g of object/5g of water = 2.4 specific gravity = (by definition) weight of object/weight of water occupying same volume = 2.4 g/cm^3.

This way, we never get Newtons involved -- just go from apparent mass straight to equivalent water.

In general, specific gravity problems on the MCAT can be handled this way -- turn everything into water, and figure from there. I believe it's fastest, and has the lowest probaility of error.