Question

A student wants to make a 0.250 M aqueous solution of silver nitrate, AgNO3, and has a bottle containing 15.89 g of silver nitrate. What should be the final volume of the solution? When you give your numerical answer, what is the correct significant figures and how do you know that is the correct amount?

Answer 1

0.37416 L

Step-by-step explanation:

M = mol/L

molar mass ( for AgNO₃ ) = 169.87 g

moles = given mass/molar mass

15.89 g/169.87 = .09354

.250 = .09354/L

= 0.37416

Answer 2

0.374 L

Step-by-step explanation:

Step 1: Given data

Mass of silver nitrate (solute): 15.89 g

Molarity of the solution: 0.250 M

Step 2: Calculate the moles of silver nitrate

The molar mass of silver nitrate is 169.87 g/mol.

`15.89g * (1mol)/(169.87g) = 0.09354mol`

Step 3: Calculate the volume of the solution

The molarity is equal to the moles of solute divided by the liters of solution.

`M = (moles\ of\ solute )/(volume\ of\ solution) \\volume\ of\ solution = (moles\ of\ solute)/(M) = (0.09354mol)/(0250mol/L) = 0.374 L`

Due to the significant figures rules, we keep 3 significant figures.