The molarity of 25.00 grams of powdered AgNO3 added to 500.00mL of pure water is 0.29 mol/L
<h3>Stoichiometry</h3>
From the question, we are to determine the molarity of the solution prepared.
First, we will determine the number of moles of AgNO₃ present.
Using the formula,
![Number\ of\ moles = \frac{Mass}{Molar\ mass}](https://tex.z-dn.net/?f=Number%5C%20of%5C%20moles%20%3D%20%5Cfrac%7BMass%7D%7BMolar%5C%20mass%7D)
From the given information,
Mass = 25.00 g
Molar mass = 169.874 g/mole
Number of moles = ![\frac{25.00}{169.874}](https://tex.z-dn.net/?f=%5Cfrac%7B25.00%7D%7B169.874%7D)
Number of moles of AgNO₃ present = 0.147168 mole
Now, for the molarity of the solution
![Molarity = \frac{Number\ of\ moles}{Volume}](https://tex.z-dn.net/?f=Molarity%20%3D%20%5Cfrac%7BNumber%5C%20of%5C%20moles%7D%7BVolume%7D)
From the given information,
Volume = 500.00 mL = 0.5 L
Molarity = ![\frac{0.147168}{0.5}](https://tex.z-dn.net/?f=%5Cfrac%7B0.147168%7D%7B0.5%7D)
Molarity = 0.294336
Molarity ≅ 0.29 mol/L
Hence, the molarity of 25.00 grams of powdered AgNO3 added to 500.00mL of pure water is 0.29 mol/L
Learn more on Stoichiometry here: brainly.com/question/14805986