# How many bromine atoms in a 246.1*g mass of bromine molecules?

There are $3.08 \cdot m o l$ $\text{bromine atoms........}$, i.e. $3.08 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ individual bromine atoms.
You simply have to know that the bromine molecule is binuclear under standard conditions, i.e. $B {r}_{2}$, and the rest of halogens are also ${X}_{2}$.
And so we gots $1.54 \cdot m o l$ with respect to molecular $B {r}_{2}$, i.e. $3.08 \cdot m o l$ bromine atoms. Are you with me?