How many mL of a 0.123 M aqueous solution of aluminum bromide, AlBr3, must be taken to obtain 7.77 grams of the salt?
Always remember, think in terms of mols.
How many mols do you want? That's gram/molar mass.
Then M = mols/L solution. You know M and mol, solve for L and convert to mL.
To find the volume of the solution needed, we can use the formula:
Volume (in mL) = (mass of substance / molar mass) / molarity
Step 1: Determine the molar mass of aluminum bromide (AlBr3).
The molar mass of AlBr3 can be calculated by adding the atomic masses of aluminum (Al) and bromine (Br), multiplied by their respective subscripts.
Aluminum (Al): 26.98 g/mol
Bromine (Br): 79.90 g/mol
Molar mass of AlBr3 = (Al atomic mass) + (Br atomic mass x 3)
Molar mass of AlBr3 = (26.98 g/mol) + (79.90 g/mol x 3)
Molar mass of AlBr3 = 26.98 g/mol + 239.70 g/mol
Molar mass of AlBr3 = 266.68 g/mol
Step 2: Substitute the given values into the formula.
Mass of substance = 7.77 g
Molar mass = 266.68 g/mol
Molarity = 0.123 M
Volume = (7.77 g / 266.68 g/mol) / 0.123 M
Volume = (0.0291 mol) / 0.123 M
Volume = 0.237 L
Since the given 0.123 M aqueous solution of aluminum bromide is in liters (L), we need to convert the result from liters to milliliters (mL):
Volume (in mL) = 0.237 L x 1000 mL/L
Volume (in mL) = 237 mL
Therefore, to obtain 7.77 grams of aluminum bromide, you would need to take 237 mL of a 0.123 M aqueous solution of aluminum bromide.