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)
9 answers
the answer is 7
no it is 7.03
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
The answer is 7
lets gooooooooooo
the answer is 7.019283
7.03
woprk is
mol/ppm= gram/ L
woprk is
mol/ppm= gram/ L
i am real indian mjahani ansywer is 7.017
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)
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)
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.
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.
3 answers