Answer:
60 g/L is the final concentration of NaI solution .
Explanation:
Molarity of NaI solution before evaporation =[tex]M_1= 0.2 M[/tex]
Volume of NaI solution before evaporation =[tex]V_1= 2.0 L[/tex]
Molarity of NaI solution after evaporation =[tex]M_2= ?[/tex]
Volume of NaI solution after evaporation =[tex]V_2= 1.0 L[/tex]
[tex]M_1V_1=M_2V_2[/tex] ( dilution)
[tex]M_2=\frac{M_1V_1}{V_2}=\frac{0.2 M\times 2.0 L}{1.0 L}=0.4 M[/tex]
Molar mass of NaI = 150 g/mol
Concentration of NaI after evaporation :
0.4 M × 150 g/mol = 60 g/L
60 g/L is the final concentration of NaI solution .
