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moles questions!!

hi i have been given some questions as homework and i am a bit stuck. if someone could help me with the first one i might understand it more.

given that the molecular weight of methylene blue is 374 what weight was dissolved in 1 litre of water to give a solution of 30 micro mol/litre?

if anyone could help i would greatly appreciate it!!!
Reply 1
Avatar for nathan00
nathan00
maybe im confused, but 30 micromoles of methylene blue isn't it?

NB one micromole is 1×10−61 \times 10^{-6} moles i think...
Reply 2
Avatar for IChem
IChem
11
n=cVn=cV

From this formula you can work out the number of mols of solute in the solution, remember that 1μl=1×10−6l1 \mu l = 1 \times 10^{-6} l

Now that you have the number of mols you can use the formula

m=nMm=nM

This should give you the weight of the solute, which should be quite a small number, of the order of 10−2g10^{-2}g.
okay all of you are confusing me! i learnt it that moles was avogrado constant which means the amount of particles in a substance as in carbon atoms in exactly 12g of the carbon 12-isotope.
the formula is
n= m/Mr or n=c*v(in cm3)/1000 or c*v(dm3)
finally, n=v(dm3)/24.0 or n=v(cm3)/24000
Reply 4
Avatar for IChem
IChem
11
Unique_Renee
okay all of you are confusing me! i learnt it that moles was avogrado constant which means the amount of particles in a substance as in carbon atoms in exactly 12g of the carbon 12-isotope.
the formula is
n= m/Mr or n=c*v(in cm3)/1000 or c*v(dm3)
finally, n=v(dm3)/24.0 or n=v(cm3)/24000



All of what you have said here is true, but mostly irrelevant, you arn't being asked to calculate how many molecules of solvent there are, merely the weight of what has been dissolved. So basically, here is a full solution, because you are getting yourself confused:

n=cVn=cV

n=(30×10−6)×1=3×10−5molsn=(30 \times 10^{-6}) \times 1 = 3 \times 10^{-5} \mathrm{mols}

Now that you have the number of mols:

n=mMr⇒m=nMr n=\frac{m}{M_r} \Rightarrow m=nM_r

m=(3×10−5)×374=0.01122gm=(3 \times 10^{-5}) \times 374 = 0.01122g

Geddit?
IChem
All of what you have said here is true, but mostly irrelevant, you arn't being asked to calculate how many molecules of solvent there are, merely the weight of what has been dissolved. So basically, here is a full solution, because you are getting yourself confused:

n=cVn=cV

n=(30×10−6)×1=3×10−5molsn=(30 \times 10^{-6}) \times 1 = 3 \times 10^{-5} \mathrm{mols}

Now that you have the number of mols:

n=mMr⇒m=nMr n=\frac{m}{M_r} \Rightarrow m=nM_r

m=(3×10−5)×374=0.01122gm=(3 \times 10^{-5}) \times 374 = 0.01122g

Geddit?


Thanks for that

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