basic, basic G. Chem question that's bothering me...

[Turkeyman]

The question is:

A 2.0 kg block had dimensions 3 cm x 5 cm x 8 cm. What is its density?

A. 1.67 x 10^-3 kg/m^3
B. 1.67 kg/m^3
C. 167 kg/m^3
D. 1.67 x 10^4 kg/m^3

I just did 3x5x8 and got 120 cm^3. 100 cm = 1 meter, so 120 cm^3 = 1.2 m^3?

From there, d = m/v --> d = 2.0 kg / 1.2 m^3

density = 1.67 kg/m^3, or choice

The book (examkrackers 1001 questions in gchem, page 9, question 108) says the answer is D. Can anyone explain this to me? Something's really wrong with the way I'm approaching this problem most likely

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[UCLAstudent]

The problem: when you convert cm to m, you have to remember that volume uses meters CUBED. So, you have to divide by 100 three times, not 1 time. Or, another way would be to convert each of the dimensions (5 cm, 3 cm, and 8 cm) to meters BEFORE multiplying. Does this make sense?

 

[Turkeyman]

The problem: when you convert cm to m, you have to remember that volume uses meters CUBED. So, you have to divide by 100 three times, not 1 time. Or, another way would be to convert each of the dimensions (5 cm, 3 cm, and 8 cm) to meters BEFORE multiplying. Does this make sense?

Ooh I was just about to post, I think I understand now -->

So basically its:

120 cm^3 * (10^-2 m/ 1 cm)^3

120 * (10^-2 m)^3

120 * 10^-6 m^3

= 1.2 * 10^-4 m^3

so now: density = 2 kg / 1.2 * 10^-4 m^3

1.67 * 10^4 kg/m^3 --> choice [D]

Thanks!

 

[Turkeyman]

too tricky! gahGWHAH!!!!!!1111111111

 

[tank you]

too tricky! gahGWHAH!!!!!!1111111111

why is it so hard to hold the shift key down until u have finished typing out all your "!"?

 

[Pontifex Maximus]

why is it so hard to hold the shift key down until u have finished typing out all your "!"?

Old internet humor, grasshopper.

 

[Turkeyman]

lol i did that on purpose!!!!111111111111

[uZ 1|\/| 4 L337 h4x0rz0rZ!!!!!11111111

 

[45408]

why is it so hard to hold the shift key down until u have finished typing out all your "!"?

Welcome to the Internet!!!!!1oneelevenahundredeleven

 

[Shrike]

Serious advice about density problems on the MCAT:

Do not just do the calculations -- you will frequently get mixed up, and in particular do things like what you did here, messing up a decimal point. Instead, do everything in terms of water.

You must know the density of water, expressed in the three obvious ways: 1000 kg (one metric ton)/m^3, 1 kg/liter, 1 g/cm^3. For this problem, use the third, obviously. There are 3 x 5 x 8 = 120 cc's in the block, so if it were water we would expect it to have a mass of 120 grams. It instead has a mass of 2kg = 2000g. 2000g/120g = ~15 (close enough for MCAT purposes) -- the substance is fifteen or so times as dense as water (i.e., its specific gravity is 15, but don't let the terminology get to you). Hence, its density, in the units given in the answer choices, is fifteen times the standard 1000kg/m^3, or 1.5 x 10^4. Only choice D is close.

This is a bit ugly and seems to involve extra steps (because it does involve extra steps). It is entirely unlike what you would do in your physics class, but this isn't your physics class. Testtakers who use this method make fewer errors on MCAT problems, and that's what matters. (Incidentally, I taught this lesson yesterday, I will teach it twice today and again later in the week, and I have done it many times before. I have seen it work. It is not the official TPR method, or likely official anyone's method, but it works.)

To recap: convert and/or compare everything to water. Know the density of water. Think of answers as multiples of what they would be for water.

A small extension of this method is appropriate for most buoyancy problems.

Good luck.

Shrike
TPR physics, verbal, bio

 

[Turkeyman]

Awesome, thanks for that tip! I'll make an effort to integrate it into my conceptual thinking, because it makes a lot of sense.

 

[Turkeyman]

Dang Shrike I'm liking that method...it makes a lot of sense, thanks a poopload =D

 

[2badr]

why is it so hard to hold the shift key down until u have finished typing out all your "!"?

 

[blz]

Another good tidbit to know for fluids at rest: every additional 10m in water is +1atm. Now you should be able to relate water to other substances. For example, what is the pressure 45meters below the surface of a fluid with a specific gravity of 2. If the substance were water it would be 4.5 atm (5.5 if atmospheric pressure as well). Because we know the SG of the fluid is 2, then we know the fluid is twice as dense as water so would have a pressure of 9atm 45 meters below (or 10atm if atmospheric pressure added). Look for proportion tricks on the PS section, this is the key to moving faster and being more accurate on this section. This is how I approached this section and usually finished it with a lot of time to spare. Because of the extra speed and accuracy, my PS scores averaged in the 12-14 range.

Just a sidenote - A good way to find tricks on the MCAT is to go back to all the problems you did mathematical work for or took a lot of time doing. Ask yourself "what could I do differently to solve this problem faster." A lot of times, there is a faster way to do things and your job is to find that method. Never assume that you must do math first!

 

[Turkeyman]

Another good tidbit to know for fluids at rest: every additional 10m in water is +1atm. Now you should be able to relate water to other substances. For example, what is the pressure 45meters below the surface of a fluid with a specific gravity of 2. If the substance were water it would be 4.5 atm (5.5 if atmospheric pressure as well). Because we know the SG of the fluid is 2, then we know the fluid is twice as dense as water so would have a pressure of 9atm 45 meters below (or 10atm if atmospheric pressure added). Look for proportion tricks on the PS section, this is the key to moving faster and being more accurate on this section. This is how I approached this section and usually finished it with a lot of time to spare. Because of the extra speed and accuracy, my PS scores averaged in the 12-14 range.

Just a sidenote - A good way to find tricks on the MCAT is to go back to all the problems you did mathematical work for or took a lot of time doing. Ask yourself "what could I do differently to solve this problem faster." A lot of times, there is a faster way to do things and your job is to find that method. Never assume that you must do math first!

Awesome! Thanks for this one too ;D

 

[Twitch]

Here is how I did it. I started reading the other explanations but they were too long. Just think back to your first year. It's fresh in my mind because I'm taking CHEM I

Density = Mass / Vol = 2.0 kg / 120 cm^3

= 1/6 * 1/10 kg/cm^3

The 1/6 gives me 0.167 so the above will be
= 0.0167 kg/cm^3; since 100 cm = 1m, cubing both sides
= 0.0167 kg/cm^3 * [(100 cm) ^3 / m^3]
= 0.0167 * 100^3 kg / m^3 = 1.67 * 100^2 kg/m^3
= 1.67 * 10^4 kg/m^3

I've really drawn out the simple math above. You can easily do it in a step or two. But considering that I'll be w/o a calc on the test, I'd rather not loose points for some trivial placement of decimal.

 

[shaggy411]

It instead has a mass of 2kg = 2000g. 2000g/120g = ~15 (close enough for MCAT purposes) -- the substance is fifteen or so times as dense as water (i.e., its specific gravity is 15, but don't let the terminology get to you). Hence, its density, in the units given in the answer choices, is fifteen times the standard 1000kg/m^3, or 1.5 x 10^4. Only choice D is close.

Hey,

Where did you get that instead it has a mass of 2kg. is "it" referring to the water? Where did you get that number from...and how would you do the same w/ith other problems when comparing to water. is 2kg a constant.. i don't think it is?

 

[psiyung]

The question is:

A 2.0 kg block had dimensions 3 cm x 5 cm x 8 cm. What is its density?

A. 1.67 x 10^-3 kg/m^3
B. 1.67 kg/m^3
C. 167 kg/m^3
D. 1.67 x 10^4 kg/m^3

I just did 3x5x8 and got 120 cm^3. 100 cm = 1 meter, so 120 cm^3 = 1.2 m^3?

From there, d = m/v --> d = 2.0 kg / 1.2 m^3

density = 1.67 kg/m^3, or choice

The book (examkrackers 1001 questions in gchem, page 9, question 108) says the answer is D. Can anyone explain this to me? Something's really wrong with the way I'm approaching this problem most likely

Understand that 1 m^3 = 1m x 1m x 1m = 100cm x 100 cm x 100 cm

so 2kg / .000120 or 2kg / 1.2x10^-4 = 1.67x10^4

that simple

 

[Turkeyman]

OMGOSH who revived my thread? lol look at the original post date hahahah.

Darn you shaggy!

This thread was very helpful though, so for those who helped me back in February I'm very thankful. It was a stupid math conversion error .

Shaggy, the 2.0kg is a given, for the weight of the block.

 

[psiyung]

OMGOSH who revived my thread? lol look at the original post date hahahah.

Darn you shaggy!

This thread was very helpful though, so for those who helped me back in February I'm very thankful. It was a stupid math conversion error .

Shaggy, the 2.0kg is a given, for the weight of the block.

LOL, sorry man. Didnt realize the thread is that old. Easy problem though

 

[Turkeyman]

Hey,

Where did you get that instead it has a mass of 2kg. is "it" referring to the water? Where did you get that number from...and how would you do the same w/ith other problems when comparing to water. is 2kg a constant.. i don't think it is?

"It" is referring to the block in the original problem. The block had a volume of 120 cm^3, which means that IF it were water, it would weigh 120 grams. 1g/cm^3 is a conversion for water as Shrike mentioned.

Next, we know that this "120 cc" block is ACTUALLY 2 kg, aka 2000 g. Sooo we calculate its density by comparing it to water's --> 2000/120 is about 15, so this means that the block's density is 15 x [density of water, to the correct units]. We know water is 1000 kg/m^3, so 15,000 kg/m^3. We can round this off to about 1.5 x 10^4 kg/m^3, and so the best answer turns out to be D.

I pretty much just re-iterated Shrike's post above...please read that over for further clarification. Back in Feb when I posted this: I took shrike's post, printed it out, and had it with me for some time until I really understood it. It helped a great deal and knocked yet another MCAT concept under the desk

 

[Turkeyman]

Yup..........This is really really really old.

Turkey,

Brings back the memories Huh???????

Just be glad you'll never be taking that horrid test again

hahahaha yes! So glad....

LOL, sorry man. Didnt realize the thread is that old. Easy problem though

hahaha no problem

I was dumb back then...and well I still am bawahah

 

[psiyung]

"It" is referring to the block in the original problem. The block had a volume of 120 cm^3, which means that IF it were water, it would weigh 120 grams. 1g/cm^3 is a conversion for water as Shrike mentioned.

Next, we know that this "120 cc" block is ACTUALLY 2 kg, aka 2000 g. Sooo we calculate its density by comparing it to water's --> 2000/120 is about 15, so this means that the block's density is 15 x [density of water, to the correct units]. We know water is 1000 kg/m^3, so 15,000 kg/m^3. We can round this off to about 1.5 x 10^4 kg/m^3, and so the best answer turns out to be D.

I pretty much just re-iterated Shrike's post above...please read that over for further clarification. Back in Feb when I posted this: I took shrike's post, printed it out, and had it with me for some time until I really understood it. It helped a great deal and knocked yet another MCAT concept under the desk

This is a great connection that Shrike made. But the question was so simple, that this kind of connection really wasnt neccessary

 

[Turkeyman]

This is a great connection that Shrike made. But the question was so simple, that this kind of connection really wasnt neccessary

Hahah yeah you're right. In fact my original problem was a stupid math error. Eeeeeee