Question

How to prepare 500cm3 of 2moldm-3 HCl solution by using 1.12gcm-3 HCl solution with 36.5% w/w percentage?

Answer 1

Well....first we calculate the molarity of the mother hydrochloric acid solution....

Explanation:

We want moles of solute per unit volume of solution...

`[HCl]="Moles of hydrochloric acid"/"Volume of solution"`

And we can work from a `1*cm^3` volume of acid solution....the which CLEARLY has a mass of `1.12*g` of which 36.5% represents `HCl`

`[HCl]-=((1.12*gxx36.5%)/(36.46*g*mol^-1))/(1*cm^3xx10^-3*L*cm^-3)=11.2*mol*L^-1`

And this is an extremely strong solution of hydrochloric acid...and there is AN IMPORTANT PRACTICAL ISSUE of SAFETY....

You add ACID to WATER, and NEVER WATER TO THE ACID.

For the final solution, `C_1V_1=C_2V_2`

By specification, `C_2=2*mol*dm^-3`; `V_2=500*cm^3`; `C_1=11.2*mol*L^-1-=11.2*mol*dm^-3`...

`V_1=(C_2V_2)/C_1-=(11.2*mol*dm^-3xx500*cm^3xx10^-3*dm^3*cm^-3)/(2.0*mol*dm^-3)`

`-=2.8*cm^3`....again it is acid to water, and never the reverse...