What volume of #38%# #"w/w"# #HCl# with #rho_"solution"=1.19*g*mL^-1# is required to prepare a #1*L# volume of #HCl(aq)# whose concentration is #0.10*mol*L^-1#?

1 Answer
Jul 24, 2017

Answer:

Approx. #8*cm^3#, i.e. #8*mL#.........

Explanation:

We first calculate the concentration of the conc. acid with respect to #mol*L^-1#.

#"Concentration"="Moles of HCl"/"Volume of solution"#

#=((1*mLxx1.19*g*mL^-1xx38%)/(36.46*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=12.40*mol*L^-1#

; this is very concentrated hydrochloric acid, and would fume at you. The conc. hydrochloric acid that we habitually use in a lab is approx. #10.5*mol*L^-1#.

Now we need #1*Lxx0.10*mol*L^-1=0.10*mol#

And we take the quotient...

#(0.10*mol)/(12.40*mol*L^-1)=8.06xx10^-3*1/L^-1=8.06xx10^-3*1/(1/L)#

#=8.1*mL#

IMPORTANT, so get it right. We always add ACID to WATER and never the REVERSE. Why not? Because if you spit in acid it spits back at you. I kid you not. And if you do work with such solutions you bear this in mind, and ALSO, YOU MUST wear safety spectacles habitually, and make sure your lab partner wears safety spectacles. You get a drop of conc. acid in your eye, that's your eye gone.