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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
1dm3 = 1000 cm3
= 50g
but we r making in 250cm3
which is 1000/250=4
so 50/4=12.5g...
hope u got it..
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)
hmmm... both make sense....but how come we divide Mr by 4..this makes it little impracticle..isnt it?