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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

Respuesta :

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 = [tex]\frac{mass}{atomic mass of one mole}[/tex]

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)

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