How do I dilute the 2mol.dm^-3 HCl?

Question

I did this practice titration experiment at school-but i'm confused about how I diluted the HCl.

"You are provided with 250 cm^3 of limewater such that it contains approx 1g per dm^3 of calcium hydroxide.
Also available is hydrochloric acid which has a concentration of exactly 2.00 moles per dm^3. This is too concentrated to be used and you will need to dilute it.
You are to plan an investigation which will allow you to determine the concentration, in grams per dm^3, of the limewater as accurately as possible."

I've worked out the balanced equation for the reaction is
Ca(OH)ˇ2 + 2HCl ---> CaClˇ2 + 2Hˇ2O
so the reacting ratio is 1:2:1:2

I know that, if theres 1g per dm^3 of Ca(OH)ˇ2, then I must find the number of moles per gm^3:

Moles=Mass÷RFM=1g÷(40+[16x2]+[1x2])=1÷74=0.135135135 moles
So theres 0.135135135moles per dm^3.
The ratio shows I need double the number of moles of HCl per dm^3 - so it'll be (1÷74)x2=0.027027207per dm^3
How do I dilute the 2 mol.dm^-3 HCl

Your answer has helped me loads Gervald F - Thanks!.
Can someone explain why my teacher would have said to use 10cm^3 of HCl in 500cm^3 water? That would made a 0.4M solution - which is way too much, right?

Answer 1

I doesn't have to be exactly 0.27...M. A reasonable concentration would be 0.0200M, and you could very easily do this by pipetting (accurately) 1.00ml of 2.00M HCl into a 100ml volumetric flask, and diluting it up to the mark. This will dilute the HCl by a factor of 100, which is what you want.