Question

How many grams of potassium chloride are required to many prepare 500.0 mL of a 0.100 M solution?

Answer 1

A bit under `4` `g` potassium chloride salt.

Explanation:

In `0.5000*L` of a `0.100*mol*L^-1` solution there are `0.5000*Lxx0.100*mol*L^-1=0.0500*mol`

Thus we require `0.0500*molxx74.55*g·mol^-1` `=` `??g" potassium chloride"`