How many atoms are present in 95.1 L #H_2O# at STP?

1 Answer
anor277
Feb 6, 2016

You should tell us. What is the state of water at STP?

Explanation:

There are 95.1 kg of water at STP; i.e #(95.1xx10^3*g)/(18.01*g*mol^-1)# #=# #??# #"moles"#.

Now the mole represents Avogadro's number of particles, #6.022xx10^23# #mol^-1#, #N_A#. So multiply the number of moles first by #N_A# to give the number water molecules, and then mulitply this quantity by #3#. Why three?

Because this is a large number, you would be justified in using the symbol #N_A# in your answer, provided you defined it.

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