# How many atoms in a 9.6*g mass of sulfur?

$\text{Number of atoms}$ $=$ $\text{Mass"/"Molar mass"xx"Avogadro's number.}$
$\text{Atoms of sulfur}$ $=$ $\frac{9.60 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.
Because the units all cancel out, the answer is clearly a number, $\cong 2 \times {10}^{23}$ as required.