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?
well then 0.4 * 150 = 60
same amount of NaI
half the water
twice the mols per liter
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
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.
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
Okay, thank you!
To find the final concentration of the NaI solution, we need to calculate the amount of NaI that remains in the solution after the evaporation process.
First, we need to determine the initial amount of NaI in the 2.0 L solution.
The concentration (C) is given as 0.2 M, which means there are 0.2 moles of NaI in 1 L of the solution.
So, in 2.0 L of the solution, the initial amount of NaI can be calculated as follows:
Initial amount = concentration x volume
Initial amount = 0.2 moles/L x 2.0 L
Initial amount = 0.4 moles
Next, we need to determine the final amount of NaI in the 1.0 L solution after the evaporation.
Since the solution has been evaporated to form a 1.0 L solution, the final amount of NaI can be calculated as:
Final amount = Initial amount x (Final volume / Initial volume)
Final amount = 0.4 moles x (1.0 L / 2.0 L)
Final amount = 0.4 moles x 0.5
Final amount = 0.2 moles
Finally, we can determine the final concentration of NaI in the 1.0 L solution by dividing the final amount of NaI by the final volume:
Final concentration = Final amount / Final volume
Final concentration = 0.2 moles / 1.0 L
Final concentration = 0.2 M
Therefore, the final concentration of the NaI solution is 0.2 M.