Question

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

Answer 1

Answer : The number of molecules present in 65 g of silver nitrate are, $2.3\times 10^{23}$

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.

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

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

As, 1 mole of silver nitrate contains $6.022\times 10^[23}$ number of molecules

So, 0.382 mole of silver nitrate contains $0.382\times (6.022\times 10^[23})=2.3\times 10^{23}$ number of molecules

Therefore, the number of molecules present in 65 g of silver nitrate are, $2.3\times 10^{23}$

Answer 2

2.3*10^23 formula units