Calculate the volume in milliliters of a 2.09 m silver nitrate solution that contains 400 g of silver nitrate agno3. round your answer to 3 significant digits.

The volume is 1.13 *10³ mL

Explanation:

Step 1: Data given

Molarity of silvernitrate = 2.09 M

Mass of silvernitrate = 400 grams

Molar mass of silvernitrate = 169.87 g/mol

Step 2: Calculate number of moles

Number of moles = mass silvernitratre / molar mass silvernitrate

Number of moles = 400.0 grams / 169.87 g/mol

Number of moles =  2.355 moles

Step 3: Calculate volume

Molarity = moles / volume

Volume = moles / molarity

Volume = 2.355 moles / 2.09 M

Volume = 1.13 L = 1130 mL = 1.13 *10³ mL

The volume is 1.13 *10³ mL