Question

What volume of a 0.0320 M `KCN` solution contains `8.50x10^-3` mol of `KCN`?

Answer 1

Approx. `270*mL`............

Explanation:

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

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

`(8.50xx10^-3*mol)/(0.0320*mol*L^-1)=0.266*1/(1/(L))=0.266*L`