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

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Answer 1 · 2017-02-04T09:37:19

$"Number of particles"$ $-=$ $"Mass"/"Molar mass"xxN_A$ , where $N_A="Avogadro's number", 6.022xx10^23$ .

I KNOW that an $18.01*g$ mass of water contains $"Avogadro's number of molecules"$ , i.e. $6.022xx10^23$ INDIVIDUAL WATER molecules.

And thus the number of water molecules, i.e. representative particles, in a $6.8*g$ mass is:

$(6.8*g)/(18.01*g*mol^-1)xx6.022xx10^23*mol^-1-="How many water molecules"$ .

How many representative particles are in 6.8 g of H_2OH_2O?