You need to make an aqueous solution of 0.223 M barium bromide for an experiment in the lab, using a 250 mL volumetric flask. How much solid barium bromide should you add?

16.558 g of BaBr2.

Explanation:

Molar concentration, M of a substance is defined as the number of moles of the solute in a unit volume of the solution.

M = 0.223 mol/l

Volume = 250 mL

Converting from mL to l,

250 ml = 0.25 l.

Number of moles = M * V

= 0.223 * 0.25

= 0.05575 mol.

Molar mass of BaBr2 = 137 + (80*2)

= 297 g/mol

Mass = molar mass * number of moles

= 297 * 0.05575

= 16.558 g of BaBr2.

16.566g

Explanation:

The 0.223M barium bromide means there are 0.223 moles of barium bromide is 1000ml of barium bromide solution.

Now, we firstly need to know the number of moles in 250ml

And that is = (0.223 * 250) / 1000 = 0.05575 mole

Now we need to know the amount in g of barium bromide to add.

From the relation : mass = number of moles * molar mass.

The molar mass of barium bromide is 297.14g/mol

mass = 297.14 * 0.05575 = 16.566g