# How many moles are present in 165 g of Manganese?

Approx. $3 \cdot \text{moles}$ of metal.
$\text{Number of Moles}$ $=$ $\text{Mass"/"Molar Mass}$
Here, $\frac{165 \cdot g}{54.94 \cdot g \cdot m o {l}^{-} 1}$ $\cong$ $3 \cdot m o l$ of manganese metal.