Question

The mass of AgNO3 is 169.9 g/mol. How many molecules of silver nitrate would there be in 25.0 g
of AgNO3 ? (1 point)
1.0 x 10 ^23 molecules
8.86 x 10 ^22 molecules
6.02 x 10 ^23 molecules
4.56 x 10 ^ 23 molecules

Answer 1

To find the number of molecules of silver nitrate (AgNO3) in 25.0 g, we can use the following steps:

1. **Calculate the number of moles of AgNO3 in 25.0 g**:
\[
\text{moles of AgNO3} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\]
Given the mass of AgNO3 = 25.0 g and the molar mass = 169.9 g/mol:
\[
\text{moles of AgNO3} = \frac{25.0 \, \text{g}}{169.9 \, \text{g/mol}} \approx 0.147 \, \text{mol}
\]

2. **Convert moles to molecules** using Avogadro's number (approximately \(6.02 \times 10^{23}\) molecules/mol):
\[
\text{molecules of AgNO3} = \text{moles of AgNO3} \times \text{Avogadro's number}
\]
\[
\text{molecules of AgNO3} \approx 0.147 \, \text{mol} \times 6.02 \times 10^{23} \, \text{molecules/mol} \approx 8.86 \times 10^{22} \, \text{molecules}
\]

So, the answer is:
\[
\boxed{8.86 \times 10^{22} \text{ molecules}}
\]