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

A bit under $4$ $g$ potassium chloride salt.
In $0.5000 \cdot L$ of a $0.100 \cdot m o l \cdot {L}^{-} 1$ solution there are $0.5000 \cdot L \times 0.100 \cdot m o l \cdot {L}^{-} 1 = 0.0500 \cdot m o l$
Thus we require 0.0500*molxx74.55*g·mol^-1 $=$ ??g" potassium chloride"