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, \(2.3\times 10^{23}\)
