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lets solve a paper ...wat do u say? after going thru the points ..... at least 2 papers today
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sure...lets solve a paper ...wat do u say? after going thru the points ..... at least 2 papers today
asha can u plz tell me y we have to divide by 4 the Mr of crystal...?When preparing a solution of a fixed concentration from a given parent solution of concentration , say, 2.0 mol/dm3
A. We are required to prepare a solution of concentration of 1.0 mol/dm3 with volume 250 cm3 ( the concn cannot be greater than 2 !!!!!!!!!!!!!!)
a. Add 100 cm3 of the parent solution to a volumetric flask of marking at 250 cm3
b. Top off with water to the mark of 250 cm3. Use a wash bottle for this purpose to have greater control over the addition process
When required to prepare 250 cm3 of 0.5 mol/dm3 solution of a crystal of Mr 50g
First realize that we need only 250 cm3, not 1 dm3
Now in 250 cm3 there will be: 50/4 = 12.5 g of the solid
So, first add 12.5 g of solid to 50 cm3 of water in a BEAKER ( not the volumetric flask yet). Also note that the volume of water is less than 250 cm3.
Stir properly and if the solid doesn't dissolve add more water until it fully dissolves
Then transfer the solution from the beaker to a volumetric flask. Remember to rinse the beaker with water and transfer the solution to the volumetric flask
Stopper the flask and shake properly
Finally, add the required volume of water to make the solution upto 250 cm3 USE A FUNNEL OR YOU RISK OVERSHOOTING THE MARK
ohhh got it...asha can u plz tell me y we have to divide by 4 the Mr of crystal...?
why... tell meohhh got it...
1dm3 = 1000 cm3why... tell me
wats concordant titres?Titration is accurate because:
1. Standard solution of acid/base is used
2. we obtain concordant titres
3. % error in pipette and burette is very small
4. The end point of a titration is sharp
never heard of these kinds of filtration ? wat r those red phrases methods? have no idea about them...For filtration:
1. If you require a great separation , use a fluted filter paper
2. If you require a quick filtration, use vacuum filtration
During magnesium and hydrochloric acid experiment, to avoid loss of gas do not start the experiment by dropping a piece of Mg in the container of acid!!!
Rather, use a divided flask (I would appreciate it greatly if someone explains what that actually means) or place a test tube containing the magnesium metal in the solution and pull the string to start the reaction
If a syringe gives wrong volume, it could be because the syringe got stuck during the experiment
Okay guys, just over with P5. Let me put down whatever I understood from the P5s of M/J 2002 to O/N 2010 into words. Hopefully someone can correct me when I make mistakes and yeah add to what I have written (of course!)
Jazzak Allahu khairan a million times over for u my friend...you've helped us ALOT
hope the conc...r correct...Suppose you are given a solution with concentration 1.0 mol/dm3 and you have to serially dilute it to various concentrations then:
Volume of solution / cm3 Volume of water/cm3 Relative concentration
50 = 0 = 1.0
40 = 10 = 0.8
30 = 20 = 0.6
20 = 30 = 0.4
10 = 40 = 0.2
5 = 45 = 0.1
I have kept the volume constant at 50 cm3 but this is not necessary. All you need to do is get the right concn using the volume of water and volume of solution. Try filling the other relative concentrations yourself.
wats divided flask? can any1 explain the red colour sentence?
what are construction lines??? and how to calculate x in a formula from it??? m/j o8 http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_5.pdf
its not m/j peper...its o/n 52...GUYS JUNE O9 VARIANT 52 how to make 250 cm3 of solution of acid???? 4 marks!!!
if graph of mass of cuo against mass lost is plotted ... still same procedure???Mr of ............ 2 CuO = 79.5*2 = 159
Mr of............. CuCo3.Cu(oH)2 = 221
Mr of............. xH20 = 18x
Mr of.............CuCo3.Cu(oH)2.xH20 = 221 + 18x
Find the gradient........ eg : 0.7
so, 159 / 221+ 18x = o.7
now simplify it.... as x= 0.34 ans.... ( ms ans is near to 0.5) based on the gradient .....
The first calculation is incorrect as someone pointed out already. The second calculation is also incorrect.When preparing a solution of a fixed concentration from a given parent solution of concentration , say, 2.0 mol/dm3
A. We are required to prepare a solution of concentration of 1.0 mol/dm3 with volume 250 cm3 ( the concn cannot be greater than 2 !!!!!!!!!!!!!!)
a. Add 100 cm3 of the parent solution to a volumetric flask of marking at 250 cm3
b. Top off with water to the mark of 250 cm3. Use a wash bottle for this purpose to have greater control over the addition process
When required to prepare 250 cm3 of 0.5 mol/dm3 solution of a crystal of Mr 50g
First realize that we need only 250 cm3, not 1 dm3
Now in 250 cm3 there will be: 50/4 = 12.5 g of the solid
So, first add 12.5 g of solid to 50 cm3 of water in a BEAKER ( not the volumetric flask yet). Also note that the volume of water is less than 250 cm3.
Stir properly and if the solid doesn't dissolve add more water until it fully dissolves
Then transfer the solution from the beaker to a volumetric flask. Remember to rinse the beaker with water and transfer the solution to the volumetric flask
Stopper the flask and shake properly
Finally, add the required volume of water to make the solution upto 250 cm3 USE A FUNNEL OR YOU RISK OVERSHOOTING THE MARK
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