A chemist adds 0.55 L of a 4.69M silver nitrate (AgNO3) olution to a reaction flask. Calculate the mass in kilograms of silver nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

0.438 kg is the mass of solute that the chemist has added to the flask

Explanation:

We have the volume and the molarity. Let's calculate the moles, that chemist added.

Volume. molarity = mol

0.55 L. 4.69 mol/L = mol → 2.58 moles

Now we can know the mass of solute (mol. molar mass)

2.58 m. 169.89 g/m = 438.3 g

Finally we convert grams to kilograms

438.3 g / 1000 = 0.438 kg

The mass of silver nitrate is 0.438 kg

Explanation:

Step 1: Data given

Volume = 0.55L

Molarity silver nitrate (AgNO3) = 4.69M

Molar mass of AgNO3 = 169.87 g/mol

Step 2: Calculate moles AgNO3

Number of moles = molarity * volume

Number of moles = 4.69 M * 0.550 L

Number of moles AgNO3 = 2.5795 moles

Step 3: Calculate mass of AgNO3

Mass AgNO3 = moles AgNO3 * Molar mass AgNO3

Mass AgNO3 = 2.5795 moles * 169.87 g/mol

Mass AgNO3 = 438.18 grams ≈ 438 grams = 0.438 kg

The mass of silver nitrate is 0.438 kg