Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

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