Question

Calculate the volume in milliliters of a 1.1 M aluminum chloride solution that contains 125.0 g of aluminum chloride. Round your answer to significant digits.

Answer 1

The volume of solution is 8.5*10^2 mL

Step-by-step explanation:

Step 1: Data given

Molarity of an aluminium chloride solution = 1.1 M

Mass of aluminium chloride = 125.0 grams

Molar mass of aluminium chloride = 133.34 g/mol

Step 2: Calculate moles AlCl3

Moles AlCl3 = mass AlCl3 / molar mass AlCl3

Moles AlCl3 = 125.0 grams / 133.34 g/mol

Moles AlCl3 = 0.9375 moles

Step 3: Calculate volume of the solution

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.9375 moles / 1.1 M

Volume = 0.852 L = 852 mL

Volume = 8.5 *10^2 mL

The volume of solution is 8.5*10^2 mL

Answer 2

852 mL

Step-by-step explanation:

Data of molarity means, moles of solute in 1L of solution

Let's convert the mass of solute, into moles

Mass / Molar mass = moles

125 g / 133.33 g/mol = 0.937 moles

In 1.1 moles of AlCl₃ are contained in 1L of solution

0.937 moles of AlCl₃ are contained in (0.937 mol . 1L) / 1.1 mol = 0.852L

Let's convert the L to mL

0.852L . 1000 mL/ 1L = 852 mL