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:

$2 L * \frac{0,2 moles NaI}{1L}* \frac{150 g NaI}{1 mol NaI}=60 g NaI$

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:

$CNaI= \frac{60 g NaI}{1 L}=60 g/L$

Have a nice day!