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^-10.090*mol*L^-1 HCl(aq)HCl(aq) compare to 0.90*mol*L^-10.90*mol*L^-1 HCl(aq)HCl(aq)?