00:01
We come here in this question.
00:03
We have different questions.
00:06
So first part of this question is calculate the concentration of an iodid solution, i -o -3 minus solution, prepared by dissolving 1 .9853 grams of kio3.
00:25
And diluting, so volume is given.
00:28
So here it is mass and volume is 513.
00:32
So we need to find out the concentration of these iodide ions.
00:38
So concentration, that is molarity.
00:44
So molarity is equal to mass divided by gram molecule weight multiplied by thousand divided by v in ml is the formula.
00:55
So here, 1 .9853 divided by molar mass of potassium iodate is 214 .14.
01:05
Grams.
01:07
So this is multiplied by thousand divided by volume is 500 ml so that is equal to 0 .01855 for m approximately and the second part is we have 25 m alquette of 0 .0195 m kio3 solution which is added to a flask containing 2 grams of ki.
01:43
2 grams of ki is given and 10 ml of 0 .5m h2s .o4 is given.
01:54
And this solution requires 34 .81 sodium, sorry, thiosulfate it is given here.
02:08
Thiosulfate.
02:11
So here, this is ml.
02:13
Volume is given.
02:15
So we need to calculate the concentration of sodium thiosulfate.
02:29
Sulfate we need to find out.
02:32
So we will see here.
02:35
The reaction is i .o3 minus plus i minus so it is five moves plus six most of h plus is it gives rise to three moves of i.
02:50
2 plus 3 more of h2 is a formula and so if you see here iodate ions concentration we can calculate here a number of most so one more of iodide ion is given so here it is two to iodine is potassium iodide is 2 grams it is given 2 grams divided by 166 that is equal to 0 .012 and this is 0 .025 multiplied by 0 .095.
03:27
So the number of most is 4 .875 multiplied by 10 power minus 4.
03:34
So one more requires 5 moles.
03:36
So here we have 0 .012 is present...