How many hydrogen atoms are present in an 18*mL18⋅mL volume of water?

1 Answer
anor277
Jul 1, 2017

A lot.........

Explanation:

We have 2xxN_A2×NA hydrogen atoms in an 18*mL18⋅mL volume of water, where N_A="Avogadro's Number"NA=Avogadro's Number, 6.022xx10^23*mol^-16.022×1023⋅mol−1. Why?

Given that rho_"water"=1*g*mL^-1ρwater=1⋅g⋅mL−1, we have a mass of 1*cancel(mL)xx1*g*cancel(mL^-1)=18*g.......

And thus a molar quantity of 2*H*mol^-1xxN_A=1.204xx10^24 "individual hydrogen ATOMS......."

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