Premed 101 Forums

# mcat question.....

## Recommended Posts

hi CAn anyone help answer this actual question (I have the answer and explaination below but I still don't understand it)

Q: One half of a Tl sample decays to Pb in 3.1 mins through the emission of beta particles. If an initially pure sample of Tl contains 7 g of lead after 9.3 mins, what was the approximate mass of the original sample?

A: Since beta particles have negligible mass, you should assume that the mass of Pb that appears is equal to the mass of Tl that decays. That means 7 g of Tl decayed. Since Tl has a half-life of 3.1 minutes, 9.3 minutes represents 3 half-lives. In three half-lives, 7/8 of the original samples decays. therefore

(7/8) (x)=7g and x= 8g (which was the mass of the original sample.

NOW my question is where did the 7/8 come from

##### Share on other sites

that emoticon blocks out the fraction7/8

I don't know how I put it there

##### Share on other sites

I think the answer comes from:

The amount of Tl decreases by 1/2 for every half-life. So, after the first half-life there is 1/2 left, after the second half-life life there is 1/2 of the 1/2 left...

so what you have remaining of the Tl is 1/2 x 1/2 x 1/2 which is 1/8. There is 1/8 of the Tl remaining, so the amount of Tl that decayed is 7/8.

##### Share on other sites

Hmm...thats intersting..I would have thought itd be 56 since..

7= (x)(1/2)^(t/3.1) where t/3.1 is the # of half lifes....Since 9.3/3.1 = 3, there are 3 half lives...so solving for X gives 56...

So if you start with 56 and divide by 2, you get 28 Pb.

Divide 28 pB by 2 and you get 14 .

Divide 14 by 2 and you get 7........But Im wondering why its 8 and not 56...Am I doing something wrong here?

##### Share on other sites

Let's give this a try. I just spent the last 8-9 hrs in the OR watching ENT stuff, so this should be right up my alley.

First, here's what we know:

1) The emitted particle is basically massless, so the total mass of the metal stays constant (It just changes from Tl to Pb).

2) We have undergone three half-lives.

So, let Z be the original mass of the Tl. From principle 1) above, Z also therefore is the mass of the Current Tl + mass of the current Pb.

Therefore, Z x 0.5 x 0.5 x 0.5 = Current Tl mass = 1/8 x Z

Therefore, since the Current Tl mass + Current Pb mass = Z

Current Pb mass = Z - 1/8 x Z = 7/8 x Z

Since Current Pb mass is given as 7 grams, then:

7 grams = 7/8 x Z

Therefore, Z = 8 grams; the original mass of Tl is 8 grams.

Ian

UBC, Med 3

##### Share on other sites

Rainnoodle,

I'm not sure about the equation that you posted (I still have to review that ). But, the question says there is 7 g of lead after 3 half-lives. By your reasoning, you would get 28g of lead after the first decay plus the 14g after the second and 7g after the last decay.

##### Share on other sites

Hi guys,

You have to be very careful about whether or not questions state that there is X grams of a substance LEFT OVER or there is X grams of substance that has DECAYED. I tutored SEC V math here in quebec (equivalent to grade 12) and this was a common mistake that people made.

Now, if there were 7 grams "left over" then the correct answer would be 56 grams originally as someone had said.

But, if we are told 7 grams had "decayed" than Ian's mathematical approach is correct.

Many of my students had a problem of figuring out what the variables were. Be careful. Be very careful... hehehe... good luck!!!

later,

tones

##### Share on other sites

Thankyou all so much!!!!!

I didn't expect such a thourough response, you guys and girls are awesome!!!!

##### Share on other sites

Hey I just found this website..... it's great! I will be writing my mcat in august and I have a few questions I would like answers for if that's okay.

here goes

1. Why does He2 not exist as a stable molecule?

2. A 700N person pushes a 900N box 5.00m up a 15.0m long ramp in 3.00 seconds. At that point on the ramp the box is 3.oom above the ground. The frictional force for the box on the ramp is 100N. What is the minimum force the person must apply to get it moving?

the answer to the question is 640N but there is no explaination, if anyone can explain it that would be a great help.

(p.s. I noticed that there are not a lot of actual MCAT questions posted so I hope that it's okay to post mine, let me know if I'm breaking a rule or something.

##### Share on other sites

For #1:

He2 does not exist because of the antibonding molecular orbitals..., with which the electrons do fill...

Recall that in a diatomic molecule, (according to the MO orbital theory), the following orbitals will be created in terms of increasing energy: a bonding sigma, antibonding sigma, and so on....but we only need to consider the 1st two orbitals, the bonding sigma, and the antibonding sigma (both at energy level=1)

Each He will have two electons (both in 1s orbital)...therefore, in a theoritical He2 molecule, there will be 4 electrons...which will fill up the a bonding sigma, and antibonding sigma orbitals...since both bonding and antibonding orbital are used, they both cancel, and you get no He2..

Compare that to an H2 molecule....each H has 1 electron (1s)....therfore two e's per H2.....these two e's will fill up only the bonding orbital...there are no more electrons left to fill the antibonding one, therefore, H2 exists....

##### Share on other sites

For question #2:

In order to get the box moving again, the person must overcome the friction force (100N) plus the gravitational force that is acting on the box down the ramp. You can calculate the gravitational force acting on the box by:

mgsinX (X (or theta) = the angle of the ramp)

you know the box weighs 900N, so it's 900sinX. sinX = 3/5 (opposite over hypotenuse). You get 900(3/5) = 540. So the person needs to overcome 540N plus the 100N of the frictional force. So you get an answer of 640N.

Did I do that right, everyone?

They just throw in all those other numbers (like the amount of time it took the person to push the box up that high) to try to throw you off.

##### Share on other sites

rainnoodle, could you expand on your explaination of the antibonding orbitals? I can't seem to find much information on them and you really seem to understand them throughly. thanks

#### Archived

This topic is now archived and is closed to further replies.

×

• Pages