Asked by Danteh

How many moles of solute are present in (3.9x10^0) g of a (2.000x10^0) ppm solution of potassium fluoride? (answer to 2 s.d in mol)
Answered by robot real
the answer is 7
Answered by robot fake
no it is 7.03
Answered by alec
first find the molar mass of potassium fluoride. Then implement the alt method in order to find the ppm(partpermillion) present. The

























The answer is 7
Answered by Dababy
lets gooooooooooo
Answered by pupkin
the answer is 7.019283
Answered by robot fake
7.03

woprk is

mol/ppm= gram/ L
Answered by jiskha (real)
i am real indian mjahani ansywer is 7.017
Answered by Mahatma
A similar unit of concentration is molality (m), which is defined as the number of moles of solute per kilogram of solvent, not per liter of solution:

molality=molessolutekilogramssolvent(15.3.1)
Mathematical manipulation of molality is the same as with molarity.

Another way to specify an amount is percentage composition by mass (or mass percentage, % m/m). It is defined as follows:

%m/m=massofsolutemassofentiresample×100%(15.3.2)
It is not uncommon to see this unit used on commercial products (Fig. 11.3.1 - Concentration in Commercial Applications)
Answered by DrBob222
Judging from what I see no knows how to work the problem. If you will show your work how obtained those values I will help you see what you did wrong.
I'm assuming this is 2.00 ppm w/w which means 2 mg KF in 1 kg solution. So in 3.9 g solution you will have 2.00 mg x 3.9/1000 = 0.0078 mg KF,
mols KF = g/molar mass =.0078E-3g/58.1 = 1.34E-7 mols KF.
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