Question #ba64c

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Question #ba64c

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Answer 1 · 2017-09-06T12:40:41

Let's underthink the problem....suppose you had $3 dozen$ $"3 dozen"$ individual water molecules, i.e. $36$ $36$ molecules......

Explanation:

Do you agree that in this scenario we gots 36 individual oxygen atoms, and 72 hydrogen atoms? Yea or nay?

The question here specified $3.4 \cdot m o l$ $3.4*mol$ water molecules, which has a mass of $3.4 \cdot m o l \times 18.01 \cdot g \cdot m o l^{- 1} = 61.2 \cdot g$ $3.4*molxx18.01*g*mol^-1=61.2*g$ of water......

But a mole is simply a numerical quantity....i.e. $N_{A} = 6.022 \times 10^{23} \cdot m o l^{- 1}$ $N_A=6.022xx10^23*mol^-1$ .

And so back to your problem, we gots....

$3.4 \times N_{A}$ $3.4xxN_A$ $oxygen atoms$ $"oxygen atoms"$ $=$ $=$ $2.05 \times 10^{24}$ $2.05xx10^24$ $oxygen atoms$ $"oxygen atoms"$ .

And......

$2 \times 3.4 \times N_{A}$ $2xx3.4xxN_A$ $hydrogen atoms$ $"hydrogen atoms"$ $=$ $=$ $4.09 \times 10^{24}$ $4.09xx10^24$ $hydrogen atoms$ $"hydrogen atoms"$ .

And the moral, the mole is simply a number, admittedly a very large number.......with the property that a mole of $hydrogen atoms$ $"hydrogen atoms"$ has a mass of $1 \cdot g$ $1*g$ .

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Question #ba64c

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