same amount of NaI
half the water
twice the mols per liter
Some amount of water is evaporated from a 2.0 L, 0.2 M NaI solution, to from a 1.0 L solution. The molar mass of NaI is 150 g/mol.
What is the final concentration of NaI solution in?
9 answers
It took a good few minutes, but I understand what you meant by the same amount of NaI and half the water, but I don't get the third. Does that mean 150 x 1.0?
Wait, I converted the 1.0 L to mL and got 1000, then divided- 150/1000 and got 15 g/L, is that correct?
The original concentration was 0.2 M
That means 0.2 mols per liter
now you still have the same mols of NaI
but half as much water
so the concentration is 0.2 mols for HALF a liter
which is 0.4 mols / liter concentration
0.40 M
That means 0.2 mols per liter
now you still have the same mols of NaI
but half as much water
so the concentration is 0.2 mols for HALF a liter
which is 0.4 mols / liter concentration
0.40 M
Concentration is normally in mols/ liter like 0.4 M
If you want grams/liter do 0.4 * 150 grams/mol
but that is not what the question asked for.
If you want grams/liter do 0.4 * 150 grams/mol
but that is not what the question asked for.
All of the answers available are # g/L, and 15 and 60 are some of them, so 60 g/L?
Mols is number of molecules times Avagadro's number (around 6*10^23)
M means mols / liter
grams/ liter = mols / liter * grams/mol = M * molecular mass
M means mols / liter
grams/ liter = mols / liter * grams/mol = M * molecular mass
well then 0.4 * 150 = 60
Okay, thank you!