I've posted a topic on Mole Calculations after completing Empirical and Molecular Forumulas, and I think this was the natural progression for me in terms of topic as it has introduced me to volumes, concentrations and solutions, etc.
I'm having to distance learn due to my circumstances, and so I'm trying my best to stick to a logical order of work for the subject based on what the exam boards will eventually expect from me.
If someone could please answer the following questions (full method shown) then that would be very much appreciated, as I've done them earlier today and want to be able to check and correct them so I know I understand the topic and can move onto the next topic by tomorrow. It's very much appreciated. Thank you.
I'm only putting it from Qu 4, as that is when they become questions with equations and so more involved.
4) Calculate the concentration of a solution of sodium hydroxide given that 25.0 cm3 of it required 18.8 cm3 of 0.0500 mol dm-3 H2SO4.
H2SO4 + 2 NaOH ® Na2SO4 + 2 H2O
5) Calculate what volume of 0.05 mol dm-3 KOH is required to neutralise 25.0 cm3 of 0.0150 mol dm-3 HNO3.
HNO3 + KOH ® KNO3 + H2O
6) A 250 cm3 solution of NaOH was prepared. 25.0 cm3 of this solution required 28.2 cm3 of 0.100 mol dm-3 HCl for neutralisation. Calculate what mass of NaOH was dissolved to make up the original 250 cm3 solution.
HCl + NaOH ® NaCl + H2O
7) What volume of 5.00 mol dm-3 HCl is required to neutralise 20.0 kg of CaCO3?
2 HCl + CaCO3 ® CaCl2 + H2O + CO2
8) 3.88 g of a monoprotic acid was dissolved in water and the solution made up to 250 cm3. 25.0 cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.
9) A 1.575 g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250 cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0 cm3 of this solution of acid reacted with exactly 15.6 cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.
10) A solution of a metal carbonate, M2CO3, was prepared by dissolving 7.46 g of the anhydrous solid in water to give 1000 cm3 of solution. 25.0 cm3 of this solution reacted with 27.0 cm3 of 0.100 mol dm-3 hydrochloric acid. Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the metal M.
11) A 1.00 g sample of limestone is allowed to react with 100 cm3 of 0.200 mol dm-3 HCl. The excess acid required 24.8 cm3 of 0.100 mol dm-3 NaOH solution. Calculate the percentage of calcium carbonate in the limestone.
12) An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm3 of 0.200 mol dm-3 hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of sodium hydroxide solution was required. 25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.
Originally posted by Tullia:I've posted a topic on Mole Calculations after completing Empirical and Molecular Forumulas, and I think this was the natural progression for me in terms of topic as it has introduced me to volumes, concentrations and solutions, etc.
I'm having to distance learn due to my circumstances, and so I'm trying my best to stick to a logical order of work for the subject based on what the exam boards will eventually expect from me.
If someone could please answer the following questions (full method shown) then that would be very much appreciated, as I've done them earlier today and want to be able to check and correct them so I know I understand the topic and can move onto the next topic by tomorrow. It's very much appreciated. Thank you.
I'm only putting it from Qu 4, as that is when they become questions with equations and so more involved.
4) Calculate the concentration of a solution of sodium hydroxide given that 25.0 cm3 of it required 18.8 cm3 of 0.0500 mol dm-3 H2SO4.
H2SO4 + 2 NaOH ® Na2SO4 + 2 H2O
5) Calculate what volume of 0.05 mol dm-3 KOH is required to neutralise 25.0 cm3 of 0.0150 mol dm-3 HNO3.
HNO3 + KOH ® KNO3 + H2O
6) A 250 cm3 solution of NaOH was prepared. 25.0 cm3 of this solution required 28.2 cm3 of 0.100 mol dm-3 HCl for neutralisation. Calculate what mass of NaOH was dissolved to make up the original 250 cm3 solution.
HCl + NaOH ® NaCl + H2O
7) What volume of 5.00 mol dm-3 HCl is required to neutralise 20.0 kg of CaCO3?
2 HCl + CaCO3 ® CaCl2 + H2O + CO2
8) 3.88 g of a monoprotic acid was dissolved in water and the solution made up to 250 cm3. 25.0 cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.
9) A 1.575 g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250 cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0 cm3 of this solution of acid reacted with exactly 15.6 cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.
10) A solution of a metal carbonate, M2CO3, was prepared by dissolving 7.46 g of the anhydrous solid in water to give 1000 cm3 of solution. 25.0 cm3 of this solution reacted with 27.0 cm3 of 0.100 mol dm-3 hydrochloric acid. Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the metal M.
11) A 1.00 g sample of limestone is allowed to react with 100 cm3 of 0.200 mol dm-3 HCl. The excess acid required 24.8 cm3 of 0.100 mol dm-3 NaOH solution. Calculate the percentage of calcium carbonate in the limestone.
12) An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm3 of 0.200 mol dm-3 hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of sodium hydroxide solution was required. 25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.
I will not do complete worked solutions for students (in other words, I won't do your homework for you), but instead will only offer comments to guide your thinking process. But even then, probably only to a few selected questions, rather than every single question asked.
But before I do any of that (that's a lot of qns you've posted, and forseeably, you'll be asking many more questions for the other topics in the next 12 months or more), I'll like to ask you to share more about yourself.
Without mentioning any info that could identify you personally (I'm not interested in your personal identity, only interested in understanding why you feel the need to ask the many questions you ask, as if you were self-studying the entire syllabus yourself and do not have at least a personal tutor or a school teacher to guide you with), so please share something of the following :
What's your approximate age? (you do not sound like a typical teenager taking 'O' levels) Which country are you from, which country are you currently studying in, and which country will you be sitting for your exams in?Why are you forced to do long-distance studying? Which exam board will be issuing your certificate? Are you taking IGCSE or GCSE or GCE, etc? Do you have, or can you afford to, or do you intend to, engage a personal tuition teacher, if you're self-studying the entire syllabus, instead of studying in a government or private school? Will you be continuing chemistry studies at 'A' levels after your 'O' levels? What course do you intend to study in the University?
You're not obliged to answer any of these questions, of course. And some might think the questions to be irrelevant. But you should be aware that, if given a better understanding your background, may result in more people being willing to reply, assist, contribute, etc, to your many questions.
But (you should be aware that) it's doubtful anyone would be willing to continually give complete worked solutions for all your questions (which forseeably might accumulate into the thousands, in the next 12 months till your 'O' level exams). That would be a hired personal tutor's job.
I respect your standpoint, and would actually prefer the guidance in answering questions more than offering worked solutions. For example, if you could choose a question of each type [like there was a limiting reactant and water of crystallisation question in the previous topic posted] and give me a guided solution for that one specific question then that would be very helpful and much appreciated.
In the previous topic, the solutions were laid out far more effectively than mine were and so I was obviously making it more difficult for myself by not having a logical and considered approach.
I'm slightly - but not considerably - older than the age expected. [Thank you, by the way. I'll take that as a compliment.] I'm from the UK, studying in the UK, and will be sitting my exams in the UK. I can't afford to go to College either full-time or part-time, and so I'm working full-time and having to complete my studies around this. AQA is the certifying exam board for my Chemistry course, which is an A-level and so considered A-level standard. [I'm not finding it difficult in terms of getting my answers correct, but I've been out of learning as I've not come straight from GCSE and so I'm not well-drilled at organising my methods and thinking logically about problems all of the time]. I intend on studying Biochemistry at University and I'm taking other A-levels via distance-learning that ensure I've got the prerequisites for entry [I could have taken other routes into University but they were more intensive in the short-term and I couldn't balance it with my worklife].
I'll just post now and again if you don't mind, and if someone wants to give some input then they can know that I'll appreciate it, but if someone doesn't wish to contribute for whatever reason then I completely understand.
Thanks for your time in reading this.
Acutally.... 99% of the question that you will be posting in future has acutally been answered by somebody else.
Search Yahoo Answer: UK & Ireland. and all the answers are there, with detail worked solution.
The power of a fingertip
Originally posted by Tullia:I respect your standpoint, and would actually prefer the guidance in answering questions more than offering worked solutions. For example, if you could choose a question of each type [like there was a limiting reactant and water of crystallisation question in the previous topic posted] and give me a guided solution for that one specific question then that would be very helpful and much appreciated.
In the previous topic, the solutions were laid out far more effectively than mine were and so I was obviously making it more difficult for myself by not having a logical and considered approach.
I'm slightly - but not considerably - older than the age expected. [Thank you, by the way. I'll take that as a compliment.] I'm from the UK, studying in the UK, and will be sitting my exams in the UK. I can't afford to go to College either full-time or part-time, and so I'm working full-time and having to complete my studies around this. AQA is the certifying exam board for my Chemistry course, which is an A-level and so considered A-level standard. [I'm not finding it difficult in terms of getting my answers correct, but I've been out of learning as I've not come straight from GCSE and so I'm not well-drilled at organising my methods and thinking logically about problems all of the time]. I intend on studying Biochemistry at University and I'm taking other A-levels via distance-learning that ensure I've got the prerequisites for entry [I could have taken other routes into University but they were more intensive in the short-term and I couldn't balance it with my worklife].
I'll just post now and again if you don't mind, and if someone wants to give some input then they can know that I'll appreciate it, but if someone doesn't wish to contribute for whatever reason then I completely understand.
Thanks for your time in reading this.
Hi Tullia, thanks for your self-intro, and yes, feel free to post your questions here every now and then.
Although here's a little tip : perhaps one or two questions at a time, might be less of a turn-off for forumners who might otherwise be willing to reply your (one or two) questions.
As I said earlier, I'll only give brief comments to guide your thinking (if anyone else reading the thead feels like contributing, don't refrain on account of my comments-only-instead-of-worked-solutions, go ahead and post the full worked solutions if you feel like it), and if my comments include any phrases you're not familiar with (eg. "molarity" or "aliquot" or "Kekule structure" or "formal charge" or "oxidation state", etc), you're advised to google and (in particular) Wikipedia them out for your own self-learning.
And it helps for forumners to know that you've skipped O levels and are studying A levels from scratch (effectively, self-studying O and A levels simultaneously). Knowing that you're ultimately taking the AQA 'A' level exam (in 12 months, I presume?) allows me to more freely use 'A' level terminology (although I personally often go somewhat beyond the 'A' level syllabuses when teaching my own 'A' level students, so be mentally prepared to do a substantial amount of googling and wikipeding when reading my comments) in my comments for you.
Firstly, remember that in chemistry calcuations, algebra is your best friend. The first couple of questions need not involve algebra, but as the questions become more involved, during your 'A' level course, you'll find increasing need for the use of algebra.
Q4. Calculate moles of H2SO4. Since it's a diprotic acid, hence calculate moles of protons H+ involved, which is also the same as the moles of hydroxide ions OH- involved, which in turn is the same as the moles of NaOH involved. Now that you've the moles of NaOH, and the Qn gives you volume of NaOH, hence determine the molarity of NaOH, commonly written as [NaOH], using the formula : Molarity = Moles / Volume (in dm3).
Q5. Calculate moles of HNO3 present. Since both acid and base are monoprotic, hence determine moles of KOH required. Since molarity of KOH is given, hence determine volume required, using the formula : Molarity = Moles / Volume (in dm3).
Q6. Calculate moles of HCl neutralized, which would be the same as moles of OH- present in the 25cm3 NaOH(aq) aliquot, hence calculate the moles of OH- present in original 250cm3 of NaOH(aq) solution (from which the aliquot was obtained), thereafter multiply by the molar mass of NaOH to determine the sample mass of NaOH required by the qn.
Q7. Let x be the volume (in dm3) of HCl required. Given the molarity of HCl, calculate moles of HCl involved, in algebraic terms. Given sample mass of CaCO3, calculate the moles of CaCO3 (simply divide the sample mass by molar mass). Bearing in mind that the CO3 2- ion is diprotic, hence calculate the moles of HCl required for acid-base neutralization. Equate this value to the moles of HCl in algebraic terms found earlier, and solve for x.
Q8. Let x be the molar mass of the monoprotic acid. Given sample mass, find moles of the acid (simply divide sample mass by molar mass), in terms of algebraic variable x. Given 250cm3 volume, determine molarity of acid in the 250cm3 solution. Since a 25cm3 aliquot was used for titration, calculate moles of H+ (still in terms of x) involved in the titration. Given molarity and volume of NaOH(aq) used in the titration, calculate moles of NaOH involved, and equate to moles of H+ found earlier (in terms of x), and solve for x.
Q9. Find molar mass of hydrated ethandioic acid, in terms of algebraic variable n. Given sample mass, find moles of acid present (in terms of n). Given volume of 250cm3, find molarity of acid in 250cm3 solution, hence determine moles of diprotic acid neutralized in titration (ie. in 25cm3 aliquot). Since ethandioic acid is diprotic, hence determine moles of H+ neutralized in titration (in terms of n). Equate this expression to the moles of OH- neutralized in titration (use the formula : moles = molarity x volume, for the NaOH), and solve for n.
Q10. Let the molar mass of metal M, be x (grams). Accordingly, determine moles of M2CO3 present, in terms of x (use the formula : moles = sample mass / molar mass; note that sample mass is given, and molar mass will be in terms of x). Given 1dm3 solution, calculate molarity (still in terms of x). Given 25cm3 aliquot involved in neutralization, determine moles of CO3 2- ions involved in neutralization (use the formula : moles = volume x molarity). Bearing in mind the carbonate(IV) ion CO3 2- is diprotic, hence calculate moles of H+ neutralized (still in terms of x). Equate this expression to the moles of HCl used (use the formula : moles = molarity x volume), and solve for x.
Q11. Let the % by mass of CaCO3(s) present in the 1g sample be x. Hence determine the sample mass of pure CaCO3(s) present, which is simply (x/100)(1g). Divide this by the molar mass of CaCO3(s) (refer to the periodic table) to obtain the moles of CaCO3(s) involved in neutralization (in terms of x). Find total moles of HCl present (moles = molarity x volume). Find moles of excess HCl by subtracting twice the moles of CaCO3, from the total moles of HCl present. (Why twice? because tghe carbonate(IV) ion CO3 2- is diprotic!). Equate this expression (of moles of excess HCl, in terms of x) with the moles of NaOH(aq) used in titration (use the formula : moles = molarity x volume), and solve for x.
Q12. Similar to Q11, but more involved and convoluted.
Let the % purity, which is the same as % by mass of Ba(OH)2 in the 1.6524 g sample, be x. Hence find sample mass of pure Ba(OH)2 present, in terms of x grams. Hence find moles of Ba(OH)2 present (formula : moles = sample mass / molar mass). Next find moles of OH- present (still in terms of x), bearing in mind that Ba(OH)2 is a diprotic base. Find total moles of HCl used (formula : moles = 0.2M molarity x 0.1dm3 volume). Subtract away moles of OH- present from total moles of HCl present, to obtain moles of HCl in excess (still in terms of x).
Given molarity of HCl, find moles of HCl present in 28.5cm3. This would give you the moles of H+ involved in the other titration, which is the same as the moles of OH- involved in this other titration. Given that in this other titration, volume of OH- used was 25cm3, hence determine molarity of NaOH(aq). Hence determine moles of OH- present in 10.9cm3. Equate this value, to the moles of HCl in excess (in terms of x) found earlier, and solve for x.
You could post all your answers here and we can check your work.
Why the need for algebra in these questions? For example, you use algebra in Q8. Isn't it simpler to just do the following:
(0.095 mol/ L) x (0.0465 L) x (1/1) = 0.0044175 mol
(3.88 g) x (25.0 / 250) = 0.388 g
( 0.388 g) / (0.0044175 mol) = 87.8 g/mol
[L notation used as it is easier to write that the British equivalent]
Sorry if I'm missing something. This is why I wanted your help. I'm humbly trying to learn.
Originally posted by Tullia:Why the need for algebra in these questions? For example, you use algebra in Q8. Isn't it simpler to just do the following:
(0.095 mol/ L) x (0.0465 L) x (1/1) = 0.0044175 mol
(3.88 g) x (25.0 / 250) = 0.388 g
( 0.388 g) / (0.0044175 mol) = 87.8 g/mol
[L notation used as it is easier to write that the British equivalent]
Sorry if I'm missing something. This is why I wanted your help. I'm humbly trying to learn.
You're not missing anything. It's just a matter of personal preference. Algebra is a tool that makes solving problems a little more wieldable, if you choose to work with it.
Nice to know. I'm really working hard in my chosen subjects at the moment and feeling I'm reaping the rewards. I'm confident I've got the correct answers for all of them now. I'm busy, but some of the more worded ones [I struggle deriving what the question is asking for at times] are towards the end of the list of questions.
For example, Q9:
(0.0156 L) x (0.160 mol NaOH/L) x (1/2) / (0.0250 L) = 0.04992 mol/L acid
(0.04992 mol/L) x (0.250 L) = 0.01248 mol acid
(0.01248 mol acid) x (90.0352 g H2C2O4/mol) = 1.1236 g acid
(1.575 g - 1.1236 g) = 0.4514 g H2O in the original sample
(0.4514 g H2O) / (18.0153 g H2O/mol) x (90.0352 g / 1.1236 g) =
2.00 mol H2O / mol H2C2O4
n = 2
Is this correct?
Yes, it's correct. Well done.