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Calculation on page CHM-108

Calculation on page CHM-108

Postby clarem4 » Wed Feb 22, 2012 6:02 am

Hi there
I am trying to understand the equation on page CHM-108 but really dont understand where the answer comes from?I understand the first step Q = It but after that I am completely lost, what is the formula Faradays = 2400C x 1 /96500C without the values substituted?Are these common questions on the GAMSAT?
Clare
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Postby goldstanda3269 » Wed Feb 22, 2012 7:18 pm

This specific problem is not a common GAMSAT question but 5-10 questions on the GAMSAT each year is dependent on your ability to pay attention to units and manipulate data with or without an equation that they may provide (i.e. dimensional analysis).

Section CHM 10.5 defines a Faraday as 96 500 coulombs and defines a coulomb as 1 amp for one second.

If you are given that the charge Q = It, given I = 2 amps and t = 20 minutes, you would be expected to know that seconds is an SI unit and not minutes.

Thus Q is determined to be 2400 C.

what is the formula Faradays = 2400C x 1 /96500C without the values substituted?


This is the point about dimensional analysis which is common on the GAMSAT: there is no equation for this, only a keen attention to the units.

Since we are looking for grams and we know grams are related to moles and moles are related to Faradays, we must convert coulombs to Faradays using the definition of Faradays (i.e. = 96 500 C).


In future, instead of starting a new thread, please post in the thread that was already created for the chapter above. Thanks for your understanding.
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