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

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 =

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)

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

it means there are 3 atoms of oxygen