# How many representative particles are in 6.8 g of H_2O?

Feb 4, 2017

$\text{Number of particles}$ $\equiv$ $\text{Mass"/"Molar mass} \times {N}_{A}$, where ${N}_{A} = \text{Avogadro's number} , 6.022 \times {10}^{23}$.
I KNOW that an $18.01 \cdot g$ mass of water contains $\text{Avogadro's number of molecules}$, i.e. $6.022 \times {10}^{23}$ INDIVIDUAL WATER molecules.
And thus the number of water molecules, i.e. representative particles, in a $6.8 \cdot g$ mass is:
$\frac{6.8 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \equiv \text{How many water molecules}$.