Chem/Orgo Book Pg. 162, Q7

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nhuebschmann
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Joined: Sun Aug 04, 2019 5:19 pm

Chem/Orgo Book Pg. 162, Q7

Post by nhuebschmann » Sat Sep 07, 2019 11:44 am

Hi there,

I was hoping someone could provide some clarification for the explanation for questions 7 from the practice passage. I've read through the work and looked back in the chapter where it talks about normality. I understand how 100 ml of RbOH was determined, but can't figure out how the mass was determined using the formula NaVa=NbVb. Also, how would we know the gram equivalent weight of RbOH is 102g/equiv? Thanks!

-Nathan
NS_Tutor_Mathias
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Joined: Sat Mar 30, 2019 8:39 pm

Re: Chem/Orgo Book Pg. 162, Q7

Post by NS_Tutor_Mathias » Fri Sep 13, 2019 12:54 am

Hey there, sorry this slipped off my radar a little bit!

You would simply know this from the molar mass of RbOh, which you can determine from the periodic table - simply the mass of Rb + O + H. This should not come from the normality equation at all.

For simplicity's sake, I suggest splitting the problem up into two parts:

1. Volume of RbOH
2. Mass of RbOH

Once you've determined the volume, multiply the volume by molarity to get the moles of RbOH you require (0.06 M * 0.1 L = 0.006 mol ). Now multiply this by the molar mass, and out pops your final answer in grams (~100*0.006 = 0.6 g). Convert that to milligrams to check with the answer choices and you're done!
nhuebschmann
Posts: 6
Joined: Sun Aug 04, 2019 5:19 pm

Re: Chem/Orgo Book Pg. 162, Q7

Post by nhuebschmann » Fri Sep 13, 2019 9:14 pm

Great, thank you!
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