What mass of potassium bromide (in grams) do you need to make 250.0 mL of a 1.50 M potassium bromide solution?

44.63g

Explanation:

First, let us calculate the number of mole of KBr in 1.50M KBr solution.

This is illustrated below:

Data obtained from the question include:

Volume of solution = 250mL = 250/1000 = 0.25L

Molarity of solution = 1.50M

Mole of solute (KBr) =.?

Molarity is simply mole of solute per unit litre of solution

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of solute (KBr) = 1.50 x 0.25

Mole of solute (KBr) = 0.375 mole

Now, we calculate the mass of KBr needed to make the solution as follow:

Molar Mass of KBr = 39 + 80 = 119g/mol

Mole of KBr = 0.375 mole

Mass of KBr =?

Mass = number of mole x molar Mass

Mass of KBr = 0.375 x 119

Mass of KBr = 44.63g

Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution

6.4 cm (max) - 25 cm (min)

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