chem 5 : Dilution

[kikie209]

how to construct the sentence for dilution..for designing experiment question
i really have no idea how to write those things..
what else do we need to describe there?
how many concentrations do we need to use?

 

[mariam]

i need help in this too..! :roll:

 

[Ahsan_8]

Say 2 M hcl is provided

Firstly measure 100 cm^3 of Hcl with the help of a measuring cylinder or a burette into a graduated (volumetric ) flask and aad eqal volume of distilled water place the stopper and shake well !! resulting solution is 1 M hcl
basically its a game of ratios ... how much water is to be used in dilution

 

[1992]

do u need to state how much water to use??
and also how u do calculate the concentration by adding water?

 

[ingenious]

dillute it using volume ratios of solution: water

The concentration can be calculated using;
(volume of solution)/(volume of solution + water) X concentration of initial solution

 

[kikie209]

oh..how bout the volume of burette or cylinders we used?
i mean do we need to state like "by using 50cm3 of burette"
ive seen some of the mark scheme requires that too...

 

[1992]

where does this formula come from?? i have never seen it :S
do we need to learn it¿¿??

 

[kikie209]

where does this formula come from?? i have never seen it :S
do we need to learn it¿¿??

we've learned it..the original formula suppose to be M1x V1 = M2x V2

 

[Buffer Solution]

graduation of the appratus has to be mentioned. And this M1V1=M2V2 formula should be used for dilution purposes.

 

[1992]

could u state waht is v1,m1, v2 and m2

 

[kikie209]

(M1 is the initial concentration of solution x V1 initial volume of solution) = (M2 is the new concetration of acid formed x V2 final volume after dilution)
u should know this equation as it is important in dilution method

 

[ammadb]

volume 1 molarity 1 volume 2 molarity 2

 

[mak]

Yeah Suppose we have 2 moldm-3 Hcl Nd u want to prepare 1 moldm-3 Hcl nd 0.5 Moldm-3 Hcl..

Take 20 cm3 of 2 moldm-3 Hcl Nd 20 Cm3 of water. now total vol is (20 + 20) = 40cm3... so Now in M1 V1 = M2 V2 we have got, (2)(20)=(40)(M2) nd we'l get the concentration of dilute solution dat is 1moldm-3... nd say we have to make all 5 differenct concentration with same total volume..
So we'll use it like.. (concentration of Solution)(vol. of acid to b used)=(total vol of solution)(desired conc. of dilute Solution)..... i.e (2)(v)=(40)(0.5)
now we'll get the vol of acid , 10cm3, nd rest 30cm3 is the water we'll add to make it 40 cm3 of 0.5 moldm-3....