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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