okay the problems are these...

Calculate the molarities of
The concentration of Na1+ in a solution that contains 3.91g of Na2S in a 1.0L of Sol'n





Describe how you would prepare 1.0L of eACH

.10M NaCl from 2.0M sol'n

.25 KClO3 from solid KIO3

2.5 M H2SO$ from solid concentrate (18M) sulfuric acid.



alright for the first problem i started like this

3.91g/1 x 1mole/78.05g = 5.01 x 10-2

5.01 x 10-2/1  x  2Na/1S

..and then im stuck.. coz im confused.


okay the last three... i have no idea how to start them i imagine that you would x amount of water to the 2.0M sol'n of NaCl but dont knw how to show that mathematically nor to figure out how much.

the second one - no clue, i guess you would have to measure the volume of a certain amout of water and then figure out of wat percent would be .25 KClO3 to that amount- but once again im clueless

the third- not a single clue.

i dont know where i got solid from- who knows i was tired- BUT we arent give the density just the problem

make a 1.0L, 2.5 M solution from a 18M concentrate.

so is the teacher in the wrong?

M1V1=M2V2

(2.5 M H2SO4)(1.0 L)  = (18 M H2SO4)(V2)
   
(2.5 M H2SO4)( 1.0 L)  = 1.388
     18 M H2SO4

   0.1388 L  x  1000mL  =  138.8 mL
1 1.0 L

Pour 138.8 mL of the sulfuric acid into a container, then add 861.2 mL of distilled water, and then there is 2.5 M of sulfric acid solution.

this is what i ended up doing... and i have a feeling its wrong as i didnt use density....

so then how else should i set it up?

Pour 138.8 mL of the sulfuric acid into a container, then add 861.2 mL of distilled water, and then there is 2.5 M of sulfric acid solution.

Sorry, brain fart. You are right.

AWK has a good point - preparation of solution is not about adding 861.2 mL, it is about filling up to 1000mL.

And second remark - it is not 138.8, rather 140. Think about significant digits.

Geodome, look what AWK wrote and you'll know why.