We need to make a #21L# volume of #0.10molL^-1# #HCl(aq)#. We start with #37.7%"(w/w)"# #HCl(aq)#, for which #rho=1.19g*mL^-1#... How is the former solution prepared?

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Answer 1 · 2018-02-08T16:54:21

An old story....

We gots conc. #HCl#..and we need to know its concentration..

We want #"moles of HCl"/"Volume of solution"#...and we work from a #1*mL# volume....

#[HCl(aq)]=((1.00*mLxx1.19*g*mL^-1xx37.7%)/(36.46*g*mol^-1))/(1.00xx10^-3*L)#

#=12.3*mol*L^-1#...and this is about as strong as you can get...

And we want....to make a #21.0*L# volume of #0.10*mol*L^-1# #HCl#, the which represents a molar quantity with respect to #HCl# of ...

#21.0*Lxx0.10*mol*L^-1=2.10*mol#

And from the conc. acid, this represents a volume of....

#(2.10*mol)/(12.3*mol*L^-1)xx1000*mL*L^-1=170.7*mL#

NOTE THAT THE ACID IS ADDED TO THE WATER AND NEVER WATER TO THE ACID....THIS IS IMPORTANT....

We need to make a #21*L# volume of #0.10*mol*L^-1# #HCl(aq)#. We start with #37.7%"(w/w)"# #HCl(aq)#, for which #rho=1.19*g*mL^-1#... How is the former solution prepared?