[dumbauburn]

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.

[AWK]

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

OK
Concentration of Na+ is two times higher

.10M NaCl from 2.0M sol'n

This is simple dilution
c1V1 = c1V2

.25 KClO3 from solid KIO3

This should be a printing error.
Weight a mass equivalent to 0.25 M KClO3 (or KIO3) and add enough water to volume of 1 L

2.5 M H2SO4 from solid concentrate (18M) sulfuric acid.

18 m H2SO4 is not solid, and you need its density

[dumbauburn]

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?

[Borek]

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?

You need density. For 18M H₂SO₄ solution it is about 1.84 g/mL.

[dumbauburn]

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?

[AWK]

Dumbauburn is absolutely right from formal poin of viewconcerning usinf dilution equation.
But we chemists (me and Borek) will use rather weight than volume for concentrated H2SO4.

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.

Dumbaubum is not right when calculate volume of water in this way. Molat solutions are prepared by bouring in a volumetric flask enough water to obtain 1 liter of solutions. For very concentrated solutions mixed with water often a final volume is less or more different from a sum of  volumes  (water + acid)

[Borek]

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.

[Donaldson Tan]

Why chemist prefer to work with weight for sulphuric acid?

[Garneck]

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