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

Answer 1

Caleb Adams

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

#"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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Answer 2

Jose Ayala

40 mL.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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