Question

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

Answer 1

Hello!

The right answer is C) 60 g/L

To know the final concentration of NaI solution, you'll need to calculate the initial amount of NaI in grams in the following way:


`2 L * (0,2 moles NaI)/(1L)* (150 g NaI)/(1 mol NaI)=60 g NaI`

Next, you only need to calculate the concentration on NaI in the new 1 L solution, because only water is evaporating from the initial solution:


`CNaI= (60 g NaI)/(1 L)=60 g/L`

Have a nice day!