How to Calculate the Volume of Aqueous Solution Needed to Obtain a Certain Mass of Aluminum Bromide

What is the formula to calculate the volume of a 0.149 M aqueous solution of aluminum bromide needed to obtain 10.1 grams of the salt?

Based on the data, how many mL of a 0.149 M aqueous solution of aluminum bromide must be taken to obtain 10.1 grams of the salt?

Calculation to Determine the Volume:

Approximately 254 mL of the 0.149 M aqueous solution of aluminum bromide must be taken to obtain 10.1 grams of the salt.

To find the volume of the 0.149 M aqueous solution of aluminum bromide needed to obtain 10.1 grams of the salt, we can use the concept of molarity (M) and the formula:

moles = mass / molar mass

First, we need to calculate the number of moles of aluminum bromide (AlBr3) present in 10.1 grams. For this, we'll need the molar mass of AlBr3.

The molar mass of aluminum bromide (AlBr3) can be calculated as follows:

Al (Aluminum) molar mass = 26.98 g/mol

Br (Bromine) molar mass = 79.90 g/mol

Molar mass of AlBr3 = (3 * Br molar mass) + Al molar mass

Molar mass of AlBr3 = (3 * 79.90 g/mol) + 26.98 g/mol

Molar mass of AlBr3 = 239.70 g/mol + 26.98 g/mol

Molar mass of AlBr3 = 266.68 g/mol

Now, we can calculate the moles of AlBr3:

moles of AlBr3 = 10.1 g / 266.68 g/mol ≈ 0.03784 mol

Next, we'll use the concept of molarity (M) to find the volume (in liters) of the 0.149 M solution that contains 0.03784 moles of AlBr3:

Molarity (M) = moles / volume (in liters)

0.149 M = 0.03784 mol / volume (in liters)

Now, let's rearrange the equation to solve for the volume:

Volume (in liters) = moles / Molarity

Volume (in liters) = 0.03784 mol / 0.149 M

Volume (in liters) ≈ 0.254 L

Finally, we'll convert the volume from liters to milliliters:

Volume (in mL) ≈ 0.254 L × 1000 mL/L

Volume (in mL) ≈ 254 mL

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