Question
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?
What is the final concentration of NaI solution in?
Answers
Damon
same amount of NaI
half the water
twice the mols per liter
half the water
twice the mols per liter
Quinn
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?
Quinn
Wait, I converted the 1.0 L to mL and got 1000, then divided- 150/1000 and got 15 g/L, is that correct?
Damon
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
Damon
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.
Quinn
All of the answers available are # g/L, and 15 and 60 are some of them, so 60 g/L?
Damon
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
Damon
well then 0.4 * 150 = 60
Quinn
Okay, thank you!