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Posted: Sun Mar 28, 2010 4:16 pm Post subject: |
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Okay so this question is really confusing... Why does the answer start with it? Shouldn't they be more specific?
Anyways so why is B wrong? I was thinking that the value they measured was lower in M, since they had a lower temperature. Hence, the recorded value of M would be LOWER than the predicted. But the answers says higher... any help? |
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admin Site Admin
Joined: 08 Dec 2003 Posts: 2176
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Posted: Tue Apr 20, 2010 10:51 am Post subject: |
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Let us clarify the solution for this question: Essentially, the ideal gas law can be used in combination with mass measurements to calculate the molar mass of an unknown gas. We can determine the amount in moles of the gas in question by PV=nRT where n=number of gas moles given all the other measured parameters P,V,R and T. Finally, the molar mass can then be computed by dividing the mass (in grams given by the problem and the stoichiometric relationship) by the calculated moles. However, the question as understood is as follows: How would the recorded value of M (molarity) compare to its predicted value with a lower temperature for the predicted value. So, we are really looking at the effect on Molarity which is n/V (mol/L) from PV=nRT at a lower temperature as this is what the temperature predicted would be.
The correct answer is A (M would be higher than expected) because of the ideal gas law relationship in which, PV=nRT and rearranged, P = (n/V)RT so that P = MRT and rearranging again gives P/M = RT so what that means is that for a lower T, M must be larger or higher due to the relationship. Moreover, T is inversely related to M and so to look at a lower T, means a higher M as a denominator of the rearranged equation. So to lower the T, M must be increased (or higher) and that is the rationale. |
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bzhou2033
Joined: 23 Aug 2010 Posts: 4
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Posted: Thu Aug 26, 2010 4:22 pm Post subject: |
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| Hi, sorry but I just wanted more clarity on this question. I don't think your response is correct. In answer explanation and in the answer to #28, M (molar mass, not molarity, as in your explanation) is clearly directly proportional to T. M = (wRT)/(PV). It even says in the explanation that as T up, M up. |
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mcat_premed3832
Joined: 19 Oct 2006 Posts: 413
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Posted: Thu Aug 26, 2010 5:09 pm Post subject: |
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Normally, "mm" stands for molar mass or you can use MW (molecular weight). However, M stands for molarity and n = the number of moles.
PV = nRT
P = (n/V)RT so that P = MRT
Thus M and T are inversely related. |
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RichardParker
Joined: 26 Nov 2010 Posts: 11
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Posted: Tue Dec 14, 2010 3:49 pm Post subject: |
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Ye this question is very confusing.
The explanation says that the predicted M should be less than recorded M.
But the answer says "It would be lower than expected." Shouldn't the recorded be higher than expected.
Unless the 'it' is referring to predicted value. All this makes it very confusing. |
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skchokshi85519
Joined: 31 Jul 2011 Posts: 2
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Posted: Sat Aug 20, 2011 8:24 pm Post subject: |
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| So, M does NOT refer to molar mass (as stated in a previous question)? |
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