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Chemo P5 tips

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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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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
asha can u plz tell me y we have to divide by 4 the Mr of crystal...?
 
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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
wats concordant titres?
 
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For filtration:
1. If you require a great separation , use a fluted filter paper
2. If you require a quick filtration, use vacuum filtration
never heard of these kinds of filtration ? wat r those red phrases methods? have no idea about them...
 
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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

wats divided flask? can any1 explain the red colour sentence?
 
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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 :D
 
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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.
hope the conc...r correct...
 
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wats divided flask? can any1 explain the red colour sentence?


first in a beaker pour HCL acid...... than take a small test tube and inside it put the Mg ribbon... tie a string on the mouth of the test tube and slowly dip the test tube by moving the string to start the reaction....
 
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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



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 .....
 
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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 .....
if graph of mass of cuo against mass lost is plotted ... still same procedure???
 
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WellWIshER said:​
GUYS JUNE O9 VARIANT 52 how to make 250 cm3 of solution of acid???? 4 marks!!!​
using a burette pour 50 cm3 of an acid into a graduated flask
fill using distilled water​
uptil blue mark​
put the lid​
and turnupside down!!​


isnt it like this?
n=c*v=30/1000*1=0.03mol/dm3
maa/g=n*Mr=0.03*90=2.7g
therefore 2.7g in 30cm3
so how much in 250cm3
cross multiply u will get 22.5g in 250cm3....
 
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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
The first calculation is incorrect as someone pointed out already. The second calculation is also incorrect.
moles required in the 250cm3 solution = .250 x .5 = 0.125
mass required = moles x Mr = .125 x 50 = 6.25g (not 12.5g)
 
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