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?
A. 30 g/L
B. 15 g/L
C. 60 g/L
D. 45 g/L
half the volume of solution with the same amount of NaI means the concentration is doubled
there are 0.4 moles of NaI in the liter of solution
sooo D?
150 g/mole * 0.4 moles/L = ? g/L
My guess it 60 g / L
*it's
To find the final concentration of the NaI solution, we need to determine the amount of solute (NaI) that remains after the water is evaporated.
We can start by calculating the initial amount of NaI in the solution:
Initial amount of NaI = Initial volume of solution × Initial concentration of NaI
Initial amount of NaI = 2.0 L × 0.2 M
Next, we need to determine the final amount of NaI remaining after the water is evaporated. Since the volume of the solution decreases to 1.0 L, we need to calculate the final concentration of NaI using the equation:
Final concentration of NaI = Final amount of NaI / Final volume of solution
To find the final amount of NaI, we will use the conservation of moles, assuming no NaI is lost during evaporation:
Initial amount of NaI = Final amount of NaI
2.0 L × 0.2 M = Final concentration of NaI × 1.0 L
Now, we can solve for the final concentration of NaI:
Final concentration of NaI = (2.0 L × 0.2 M) / 1.0 L
Final concentration of NaI = 0.4 M
However, the question asks for the concentration in g/L, not mol/L. To convert from molarity to grams per liter, we need to multiply the final concentration by the molar mass of NaI:
Final concentration of NaI in g/L = Final concentration of NaI × Molar mass of NaI
Final concentration of NaI in g/L = 0.4 M × 150 g/mol
Final concentration of NaI in g/L = 60 g/L
Therefore, the final concentration of the NaI solution is 60 g/L, which corresponds to option C.