Question

What volume of a 0.40 M `Mg(NO_3)_2` solution contains 95 g of `Mg(NO_3)_2`?

Answer 1

Approx. `1.60*L`

Explanation:

`"Concentration"` `=` `"Moles of solute"/"Volume of solution"`

And thus `"Volume of solution"` `=` `"Moles of solute"/"Concentration"`

`=` `((95*g)/(148.32*g*mol^-1))/(0.40*mol*L^-1)~=??L`