Question

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

Answer 1

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`