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Eduardwww [97]
2 years ago
8

What is the molarity of 25.00 grams of powdered (solid) AgNO3 added to 500.00mL of pure water. The molar mass of AgNO3 is 169.87

4g/mole or also written as 1mole/169.874g
1 Calculate the moles of AgNO3 added.

2 Divide by total volume. M= n/v

3 Ensure your units are correct (that means use your dimensional analysis)

Here are your hints.

The moles of AgNO3 is between 0.100 and 0.500

The numerical value of your concentration should be between 0.100 and 0.500
Chemistry
1 answer:
nlexa [21]2 years ago
4 0

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}

From the given information,

Mass = 25.00 g

Molar mass = 169.874 g/mole

Number of moles = \frac{25.00}{169.874}

Number of moles of AgNO₃ present = 0.147168 mole

Now, for the molarity of the solution

Molarity = \frac{Number\ of\ moles}{Volume}

From the given information,

Volume = 500.00 mL = 0.5 L

Molarity = \frac{0.147168}{0.5}

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

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