If 10 grams of AgNO3 is available, what volume of 0.25 M AgNO3 solution can be prepared?

1 Answer
anor277
Sep 20, 2016

Approx. #0.25*L#, but you should make a better approximation

Explanation:

#"Moles of silver nitrate"# #=# #(10*g)/(169.87*g*mol^-1)# #=# #0.0589*mol#.

We require #0.25*mol*L^-1# #AgNO_3(aq)#, i.e.

#(0.0589*mol)/"Volume of solution"# #=# #0.25*mol*L^-1#

Or #"Volume of solution"# #=# #(0.0589*cancel(mol))/(0.25*cancel(mol)*L^-1)# #~=# #0.25*L#

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