Read the statement.

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?

30 g/L
15 g/L
60 g/L
45 g/L

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

Explanation:

Molarity of NaI solution before evaporation =

Volume of NaI solution before evaporation =

Molarity of NaI solution after evaporation =

Volume of NaI solution after evaporation =

( dilution)

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 .

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the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.