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sarah1004 · Nov 6, 2009, 06:17 PM

(1L)(1.25mol/0.35L)(40g/1mol)=142.86g

I really hope This is corret :p:confused::eek:

Is it correct?

Thank you~!~!~

Unknown008 · Nov 6, 2009, 08:09 PM

Well, I never do these on a line...

Ok, here's how I do it:

350 mL of 1.25 M = (1.25*350)/1000 = 0.4375 mol
0.4375 mol -> 0.4375*40 = 17.5 g

Ok, if I examine your work, you should multiply volume and concentration and mass. So, $0.35L\ \times \ \frac{1.25mol}{1L}\ \times\ \frac{40g}{1mol}$

pramod awasthi · Nov 8, 2009, 10:38 AM

OK you want to know gramwt of NaOH for 1.25M concentration of 35 mL volume
OK
M=1.25
molecular wt=40
formula is
M=gmwt*acidity/molecularwt*volume in L
so GMWT=M*molecularwt*volume in mL/acidity*1000

1.25*40*350/1*1000
=17.50 gm requaired I confidentially

pramod awasthi · Nov 8, 2009, 10:40 AM

:p

Originally Posted by Unknown008

Well, I never do these on a line...

Ok, here's how I do it:

350 mL of 1.25 M = (1.25*350)/1000 = 0.4375 mol
0.4375 mol -> 0.4375*40 = 17.5 g

Ok, if I examine your work, you should multiply volume and concentration and mass. So, $0.35L\ \times \ \frac{1.25mol}{1L}\ \times\ \frac{40g}{1mol}$

sarah1004 · Nov 8, 2009, 01:47 PM

Yes , so answer is 17.5g right?

Thanks