How does the concentration of #0.090molL^-1# #HCl(aq)# compare to #0.90molL^-1# #HCl(aq)#?

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Answer 1 · 2016-03-27T17:01:55

Is an order of magnitude LESS concentrated.

We simply divide the second concentration by the first:

#(0.90*mol*L^-1)/(0.09*mol*L^-1)# #=# #10#.

How does the concentration of #0.090*mol*L^-1# #HCl(aq)# compare to #0.90*mol*L^-1# #HCl(aq)#?