How many molecules are there in 65 g of silver nitrate #AgNO_3#?

1 Answer
anor277
Apr 4, 2016

The molar quantity of silver nitrate is #(65.9*g)/(169.87*g*mol^-1)#

Explanation:

There are thus #(65.9*g)/(169.87*g*mol^-1)# #xx# #N_A# individual formula units of #AgNO_3#, where #N_A ="Avogadro's number",# #6.022140857xx10^23*mol^-1#.

Thus there are #(65.9*g)/(169.87*g*mol^-1)# #xx# #N_A# silver atoms, the same number of nitrogen atoms. and #(65.9*g)/(169.87*g*mol^-1)# #xx# #3xxN_A# oxygen atoms, because of the consitutional makeup of #AgNO_3#.

Capisce?

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