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Topic: 1 Mole question  (Read 3239 times)

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xian

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1 Mole question
« on: August 06, 2007, 04:38:18 AM »

Is that correct?

2. 3.Yo u ne.ed a 10mM solution of a p.otent (and dangerous) transcriptio.n inhibitor (mol wt 500) for experiment. To avoid handling the po.wder form of this c.ompound you decide to make up the whole bottle rather than attempt to weigh out a portion. The compound comes in 200mg batches. What volume would you add to one of these bottles to make up a 10mM solution?

My work so far: 10mM = 10mmol/L

Moles = mass/molarmass

So we have 200mg/500 = 2.5mmols (millimoles)

Not sure where to go from here.

Thanks for the help
« Last Edit: August 06, 2007, 09:01:40 AM by xian »

AWK

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Re: 1 Mole question
« Reply #1 on: August 06, 2007, 09:07:00 AM »
Quote
1. Phos phoric acid (H3PO4) is supplied as a concentrated liquid.  A bottle was assayed and found to have a density of 1.61g/ml and a purity of 91.4%.  The mol. wt of H3PO4 is 98.  What is the molarity of the solution.

Ok Im not looking for a straight answer, some hints/working would be greatly appreciated.

So molarity is moles of solute per litre of solution.  So would the answer just be 1.61g/ml * 0.913 (purity) which gives us 1.47g/ml, which is 1470g/L.

Is that correct?

Use purity as scale factor to mass content of H3PO4 in 1 liter of solution (or to percentage content), not to density.
AWK

xian

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Re: 1 Mole question
« Reply #2 on: August 06, 2007, 09:09:11 AM »
Thanks, yep just figured that out before you posted.  I should have been considering it in 1 litre of solution.

Still unsure on the other qu.

Dan

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Re: 1 Mole question
« Reply #3 on: August 06, 2007, 11:19:16 AM »
So we have 200mg/500 = 2.5mmols (millimoles)

Calculator says no.

200/500 = 0.4

Once you have determined the correct number of mmol in 200mg of your sample, you need to use the relation:

Concentration (mM) = Amount (mmol)/Volume (L)

And solve for Volume (since you know Concentration and Amount).

My research: Google Scholar and Researchgate

xian

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Re: 1 Mole question
« Reply #4 on: August 07, 2007, 10:03:53 AM »
Thanks, worked out the correct answer - 0.04L.