Question

Read the statement.

Answer 1

Answer:

60 g/L is the final concentration of NaI solution .

Explanation:

Molarity of NaI solution before evaporation =$M_1= 0.2 M$

Volume of NaI solution before evaporation =$V_1= 2.0 L$

Molarity of NaI solution after evaporation =$M_2= ?$

Volume of NaI solution after evaporation =$V_2= 1.0 L$

$M_1V_1=M_2V_2$ ( dilution)

$M_2=\frac{M_1V_1}{V_2}=\frac{0.2 M\times 2.0 L}{1.0 L}=0.4 M$

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 .