# How many moles are contained in 22.23 g KBr?

Approx. $0.20$ $m o l$.
$\text{No. of moles}$ $=$ $\text{Mass"/"Molar Mass}$ $=$ $\frac{22.23 \cdot \cancel{g}}{119.00 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ ??*mol.