00:01Welcome to class. Fill on the calculations in chemistry.
00:08So you have a question under solubility.
00:14So under solubility and you have a question like this.
00:21Ok, we have a question. We saw the previous one before.
00:25So if you are just joining us, you can please like, subscribe to this channel.
00:30Then refer to the previous videos on the calculations you can still have done.
00:33So now we continue with question 3 on that, solubility.
00:37So you have a question like this.
00:39Water was added to 80 grams of a salt and the salt is SEL to produce 30 CMQ of saturated
00:48solution at 25 degrees Celsius.
00:50The solubility of the salt at 25 degrees Celsius is 8 moles per GMQ.
00:55Calculate the mass of the salt that remains on the salt.
00:59So this is the question.
01:02Yes ma'am.
01:04Ok, alright, thank you sir.
01:12Yes, so you have a question like this.
01:16Now we are calculating the mass of the salt that remains on dissolved.
01:20The mass, you pay attention to this term, on dissolved, on dissolved.
01:27So if we are solving this, the solution is this.
01:31First, we get the, we are going to calculate the molar mass of this salt that is represented as SCl2.
01:42So this S chloride salt, so we say molar mass, molar mass of SCl2 is equal to, so we have
01:57been given with the total,
01:58because if we check the question here, they mention that, let's do atomic mass of S, which is the, one
02:04of the elements,
02:05it is an unknown element, but the representative of S, is 29.
02:09So you are going to use it that way, so you say 29.
02:12And how many moles do we have here?
02:14We have one, so you say times one.
02:16So plus, then chlorine is given to us as 35.5.
02:24Then how many atoms of it do we have?
02:27We have two, so it's 35.5 times two.
02:33So this becomes 29 plus 35.5 times two is 71.
02:46This is 71.
02:48Now the, that amount of 29 plus 71 gives us 100 grams, 100 grams.
02:58Now 100 grams, so this gives us 100 grams.
03:02Now we have known the molar mass now.
03:05So the next step is the mole.
03:08They told us that the mole is 8 moles per dmq, meaning that it was the solubility of the salt
03:16and that this temperature of the prime division is 8 moles per dmq.
03:22So 8 moles in 1 dmq make the saturated solution.
03:26So 8 moles, 8 moles in 1 dmq makes a saturated solution.
03:46So this means that, so we multiply the moles now times the molar mass.
03:53So that means that 8 moles of SPL2, which is 100 grams, equal to 100 grams, 8 moles of SPL2,
04:23which is 100 grams, equal to 100 grams.
04:23So it's 100 grams, gives 800 grams, gives 800 grams, gives 800 grams, gives 800 grams, in 1000 cm3.
04:52So that means that, so that means that, for every 8 moles, for, it means that 800 grams of, 800
05:06grams of,
05:09SCL2 dissolves in 1000 cm3.
05:18So therefore, if 800 grams of SCL2 dissolves in 1000 cm3, then from the question, we see 30 cm3.
05:31So what will 30 cm3, so what will 30 cm3 dissolve in?
05:35So 30 cm3, the mass that will dissolve in 30 cm3, or let's write it this way so it will
05:44not be confusing.
05:45We say 100 cm3, we say 100 cm3 requires 800.
05:54So we say 1000, we can simply say 1000 cm3.
06:071000 cm3 requires 800 grams of SCL2.
06:22So 30 cm3.
06:25So 30 cm3, we require 30 times 800, times 800, times 800 over, times 800 over, times 800 over X.
06:45So here we are doing eight.
07:02So one thousand cm3.
07:11requires 800 grams of LCl2.
07:19So therefore, 30ClQ will require 30 times,
07:2930CMQ times 800G over 1000CMQ.
07:39So this 20 take away this 20,
07:44this one take away this.
07:46So which is now 3 times 8,
07:513 times 8 is 24.
07:53So that is 24 grams.
08:04Now, 24 grams is what is required to dissolve in 30CMQ.
08:17Now, if there is 30CMQ,
08:20the undissolved salt,
08:23which is what we have to apply to,
08:25the mass of undissolved salt,
08:28mass of undissolved salt
08:36is equal to,
08:38what I was adding to 80 grams of salt.
08:40So 80 grams, 80 grams minus what we have here,
08:49which is the one that was initially required is 24 grams,
08:55will now give us 80 minus 24,
09:00that gives us 56 grams.
09:04So the mass that is undissolved is 56 grams.
09:10So mass of undissolved salt is 56 grams.
09:13So this is how we solve for undissolved salt.
09:23So first, let's quickly run through it again.
09:26So first, we calculate the molar mass of the salt
09:30that is given to us,
09:31which is SCl2.
09:33Now, 29 is the mass that was given
09:37the relative atomic mass of the S.
09:38Then we have one atom of it,
09:40so times 1,
09:41which is 29.
09:43Then 35 is for chlorine.
09:45So 35.5 times 2 gives us 71.
09:48So addition of 2,
09:50summation of the 2 gives us 100 grams.
09:53Now, 8 moles in 1 DNQ makes a saturated solution.
09:57Now, the 8 moles of 100 ground,
10:01that is 8 moles of underground
10:12of underground SCl2
10:18gives 800 ground in 1,000 CMQ.
10:22That is 1 DNQ.
10:23Because they said 1 DNQ.
10:25GMQ is 1,000 CMQ.
10:30So 1 DNQ is equivalent to 1,000 CMQ.
10:33So 1,000 CMQ requires 800 ground.
10:37That is what it means.
10:38So for that 8 moles,
10:40it means 800 ground,
10:431,000 CMQ requires 800 ground.
10:45Now, 30 TNQ therefore requires
10:4830 times we cross-multiply.
10:5030 times 800 divided by 1,000.
10:53This gives us 24.
10:54So we subtract this from the mass that was added.
10:59So 80 minus 24,
11:01which gives us 26.
11:02This is what you see,
11:03is the mass of the undissolved salt.
11:07So, thank you for watching.
11:09Please like, subscribe and share.
11:12Share, like and subscribe.
11:15Thanks for watching.
Comments