Question

How many molecules are there in 65 g of silver nitrate (AgNO3)?

Answer 1

2.3*10^23 formula units

Answer 2

Answer : The number of molecules present in 65 g of silver nitrate are,
`2.3* 10^(23)`

Solution : Given,

Mass of silver nitrate = 65 g

Molar mass of silver nitrate = 169.87 g/mole

First we have to calculate the moles of silver nitrate.

`\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=(65g)/(169.87g/mole)=0.382mole`

Now we have to calculate the number of molecules present in silver nitrate.

As, 1 mole of silver nitrate contains
`6.022* 10^[23}` number of molecules

So, 0.382 mole of silver nitrate contains
`0.382* (6.022* 10^[23})=2.3* 10^(23)` number of molecules

Therefore, the number of molecules present in 65 g of silver nitrate are,
`2.3* 10^(23)`