How many atoms in a $25.0 \cdot g$ $25.0*g$ mass of water?

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How many atoms in a $25.0 \cdot g$ $25.0*g$ mass of water?

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Answer 1 · 2016-03-28T21:22:39

We calculate the number of moles of water, and then multiply this number by $3 N_{A}$ $3N_A$

Explanation:

$Moles of water$ $"Moles of water"$ $=$ $=$ $\frac{25.0 \cdot g}{18.01 g \cdot m o l^{- 1}}$ $(25.0*g)/("18.01 "g*mol^-1)$ $=$ $=$ $1.39 \cdot m o l .$ $1.39*mol.$

In each mole of water there are $3$ $3$ $m o l$ $mol$ atoms.

So $Number of atoms$ $"Number of atoms"$ $=$ $=$ $1.39 \cdot m o l \times N_{A} \times 3 atoms m o l^{- 1}$ $1.39*molxxN_Axx3" atoms "mol^-1$ $=$ $=$ $4.16 \times N_{A} atoms$ $4.16xxN_A " atoms"$ .

$N_{A} = Avogadro's number$ $N_A = "Avogadro's number"$ $=$ $=$ $6.022 \times 10^{23}$ $6.022xx10^23$

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How many atoms in a $25.0 \cdot g$ $25.0*g$ mass of water?

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