Question

Read the statement.

Answer 1

the final concentration of NaI solution in 60 grams/litre.

Explanation:

Given that:

Initial concentration of NaI solution M1 = 0.2 M

initial volume of NaI  V1 = 2 L

Final volume  V2 = 1 Litre

Final molarity=?

concentration in grams/litre = ?

molar mass of NaI = 150 gram/mole

For dilution following formula is used:

M1 V1 = M2V2

putting the values in the equation

0.2 X 2 = 1 X M2

M2 = 0.4

For concentration in grams/litre, formula used

molarity = $\frac{mass}{atomic mass of one mole}$

mass = 0.4 x 150

          = 60 grams

So, 60 grams of NaI will be present in final solution of NaI after evaporation.

The concentration is 60 grams/ L (as volume got reduced to 1 litre from 2 litres)