Question

What is the molarity of 10.0 g of `KCI` in 0.500 L of solution?

Answer 1

`"Molarity"` `approx 0.25*mol*L^-1` with respect to potassium chloride.

Explanation:

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

`=` `(10.0*g)/(74.55*g*mol^-1)xx1/(0.500*L)` `=` `??*mol*L^-1.`