Question

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.

Answer 1

Answer:

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

Answer 2

Answer: The volume of solution is 1.13 L

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Molarity of solution = 2.09 M

Given mass of silver nitrate = 400. g

Molar mass of silver nitrate = 170 g/mol

Putting values in above equation, we get:

$2.09M=\frac{400}{170\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{400}{170\times 2.09}=1.13L$

Hence, the volume of solution is 1.13 L