Originally posted by ultimatenolifer:1) 5 students each dissolved an indegestion tablet in 100cm3 of water. They then titrated 25cm3 of their solns with dil hcl, using the same indicator. The results are shown below
student 1- 20.4 (titration value)
student 2- 20.5
student 3- 20.4
student 4- 20.6
student 5- 22.0
Which statement explains the result obtained by student 5?
a) the burette was washed out with hcl
b) the pipette was washed out with tablet soln
c) the student measured the top of meniscus in the pipette
d) the titration flask was washed out with tablet soln
ans is d) but why?
2) Which soln contains the greatest conc of hydrogen ions?
a) 2mol/dm3 of h2co3
b) 2mol/dm3 of ch3cooh
c) 2mol/dm3 of h3po4
d) 2mol/dm3 of h2so4
my ans is c) but ans given is d)
3) During the electrolysis of an aq soln of a molybdenum salt, 48g of molydenum metal (proton no 42) is deposited at the cathode by 2 moles of electrons. The formula of the molybdenum ion is probably
a) Mo4+
b) Mo3+
c) Mo2+
d) Mo+
ans given is a) but why
4) The same quantity of electricity is apssed thru 2 solns, one containing dil h2so4 and the other an aq soln of salt U2(s04)3. If the vol of h2 gas liberated under room conditions is 2.4dm3, the number of moles of U deposited is
a) 0.0333
b) 0.0667
c) 0.100
d)0.300
ans given is b) but why
Q1. Since more H+ is needed during the titration, there must have been more base (either OH- or CO3 2-, depending on the indigestion tablet) present in the titration flask.
Q2. As Darkness Hacker said, sulfuric acid is a stronger acid than phosphoric acid. Nonetheless, not a fair qn to be asked at 'O' levels, and Cambridge wouldn't ask it.
Q3. From sample mass and molar mass, find moles of metal. Since 2 moles of electrons are required to reduce so much of metal ions to metal atoms, you can work out the moles of electrons required to reduce 1 mole of metal ions, and hence deduce the oxidation state or ionic charge of the metal ion.
Q4. From volume of H2 gas, find moles of H2 gas, hence find moles of electrons transfered. From formula U2(SO4)3, you know the metal ion is tripositive U3+. Hence based on moles of electron transfered, calculate moles of metal atoms deposited.
1) A sample of 0.10mol of a metal X (Ar=27) was burned in oxygen. The resulting oxide had a mass of 5.1g. How many moles of oxygen have combined with 0.10mol of metal X?
a) 0.10
b) 0.15
c) 1.5
d) 2.7
Ans is b) but i calculated it to be 0.075
2) Why is this eqn 2kno3 => 2kno3 + o2 considered a redox rxn?
3) A graph is shown (v hard to plot, i will try my best to explain here). Y axis is solubility while x axis is temperature. There is a temp point T indicated at x axis. Below temp T, salt G has lower solubility than salt H. Above temp T, salt G has higher solubility than H. Both salts G and H solubility increase as temp increases.
a) both salts are best recovered by crystallisation
b) both salts are recovered by cooling to T temperature
c) salt of H by crystallisation and G by evaporation
d) salt of G by crystallisation and H by evaporation
Ans is d
4) If molten sodium is used as the electrolyte, why is there no product formed at the electrodes?
5) Nickel (ii) hydroxide reacts with dil H2S04 according to the following eqn
Ni(OH)2 + H2SO4 => NiSO4 + H2O
Which of the following procedures would be used in order to obtain a pure sample of the nickel (ii) sulfate?
a) adding excess dil H2So4
b) adding excess nickel (ii) hydroxide
c) methyl orange is added to indicate the end of the rxn
d) react equal masses of nickel (ii) hydroxide and dil h2so4
Ans is b
Originally posted by ultimatenolifer:1) A sample of 0.10mol of a metal X (Ar=27) was burned in oxygen. The resulting oxide had a mass of 5.1g. How many moles of oxygen have combined with 0.10mol of metal X?
a) 0.10
b) 0.15
c) 1.5
d) 2.7
Ans is b) but i calculated it to be 0.075
2) Why is this eqn 2kno3 => 2kno3 + o2 considered a redox rxn?
3) A graph is shown (v hard to plot, i will try my best to explain here). Y axis is solubility while x axis is temperature. There is a temp point T indicated at x axis. Below temp T, salt G has lower solubility than salt H. Above temp T, salt G has higher solubility than H. Both salts G and H solubility increase as temp increases.
a) both salts are best recovered by crystallisation
b) both salts are recovered by cooling to T temperature
c) salt of H by crystallisation and G by evaporation
d) salt of G by crystallisation and H by evaporation
Ans is d
4) If molten sodium is used as the electrolyte, why is there no product formed at the electrodes?
5) Nickel (ii) hydroxide reacts with dil H2S04 according to the following eqn
Ni(OH)2 + H2SO4 => NiSO4 + H2O
Which of the following procedures would be used in order to obtain a pure sample of the nickel (ii) sulfate?
a) adding excess dil H2So4
b) adding excess nickel (ii) hydroxide
c) methyl orange is added to indicate the end of the rxn
d) react equal masses of nickel (ii) hydroxide and dil h2so4
Ans is b
As mentioned in a previous post, I'll leave 'O' level Chem qns for others to help out.
Nonetheless, a couple of comments :
Q1. "Moles of oxygen" is ambiguous, as it could refer either to O2 or to O.
Q2. Question is erroneous, or has a typo.
Q3. Question is erroneous, ionic salts cannot be obtained by evaporation. Furthermore, even if they have different solubilities at different temperatures, you cannot completely separate them via crystalization or 'evaporation', even with repeated processing.
Q4. Because there isn't a cation available to be reduced at the cathode, nor an anion available to be oxidized at the anode. You'll need an ionic electrolyte for electrolysis.
Q5. The missing info necessary to answer the question is : nickel(II) hydroxide is insoluble.
Thx ultimaonline for your prompt replies. Qn5 did not indicate the state for nickel hydroxide. May i know how would this impact the answer?
By the way, o lvl chem mcqs are approaching soon. I will be attempting many school papers within these 2 days. Most probably, i will post all my accumulated mcq queries tomorrow nite. Hope whoever that reads this thread can give some inputs asap (coz running out of time haha). Anyway thanks a lot guys, esp ultimaonline for these 2 yrs of prompt accurate replies. I appreciate your help.
Originally posted by ultimatenolifer:Thx ultimaonline for your prompt replies. Qn5 did not indicate the state for nickel hydroxide. May i know how would this impact the answer?
By the way, o lvl chem mcqs are approaching soon. I will be attempting many school papers within these 2 days. Most probably, i will post all my accumulated mcq queries tomorrow nite. Hope whoever that reads this thread can give some inputs asap (coz running out of time haha). Anyway thanks a lot guys, esp ultimaonline for these 2 yrs of prompt accurate replies. I appreciate your help.
No problem. Feel free to continue posting your questions here when you're in JC next year (be sure to take H2 Chemistry!).
Since Ni(OH)2 is solid, you can add excess of it and remove it via simple filtration. In contrast, you cannot add excess H2SO4, because it will contaminate the salt crystals.
Hi ultimaonline, what do u mean by ''contaminating salt crystals?'' Do u mean the crystals will be acidified? And why is c) methyl orange is added to indicate the end of the rxn not acceptable? Because i thought by ensuring endpoint, we can be sure that neutralisation takes place and thus, pure crystals are formed?
hopefully, my last 2 qns on this thread :)
1) Is it a requirement for substances to be separated by paper chromatography to be liquids? If not, any eg. of solid substances that can be tested?
2) In an expt, 4.0cm3 of 1.0mol/dm3 aq iron(iii) chloride was mixed with 4.0cm3 of 1.0mol/dm3 aq NaOH.
Fecl3(aq) + 3NaOH(aq) => Fe(OH)3(s) + 3Nacl(aq)
What would be observed when the rxn was complete?
a) Brown ppt in a green soln
b) Green ppt
c) Brown ppt in yellow soln
d) Green ppt in pale yellow soln
Originally posted by ultimatenolifer:Hi ultimaonline, what do u mean by ''contaminating salt crystals?'' Do u mean the crystals will be acidified? And why is c) methyl orange is added to indicate the end of the rxn not acceptable? Because i thought by ensuring endpoint, we can be sure that neutralisation takes place and thus, pure crystals are formed?
It means the crystals will be a mixture of sulfuric acid crystals and nickel sulfate crystals.
Methyl orange will contaminate the crystals, just as excess sulfuric acid would (see preceeding statement).
You have to redo the titration without indicator, once you've determined the correct volume of acid to be added in the preliminary titration (ie. with indicator).
Originally posted by ultimatenolifer:hopefully, my last 2 qns on this thread :)
1) Is it a requirement for substances to be separated by paper chromatography to be liquids? If not, any eg. of solid substances that can be tested?
2) In an expt, 4.0cm3 of 1.0mol/dm3 aq iron(iii) chloride was mixed with 4.0cm3 of 1.0mol/dm3 aq NaOH.
Fecl3(aq) + 3NaOH(aq) => Fe(OH)3(s) + 3Nacl(aq)
What would be observed when the rxn was complete?
a) Brown ppt in a green soln
b) Green ppt
c) Brown ppt in yellow soln
d) Green ppt in pale yellow soln
The substance to be tested via paper chromatography need to be soluble in the solvent used (eg. water, ethanol, etc). Solids will not travel up the chromatograph paper.
NaOH is limiting, and Fe3+(aq) is in excess. Hence, the product is a brown ppt of Fe(OH)3 in excess Fe3+(aq) which is a pale yellow solution.
Noted with thanks. However, is it possible for the substance to be a solid and it just needs to be soluble in the solvent?
Originally posted by ultimatenolifer:Noted with thanks. However, is it possible for the substance to be a solid and it just needs to be soluble in the solvent?
Yes, but by definition, when a solid dissolves in any solvent (not just water), it is no longer a solid!