what mass of KBr (in grams) do you need to make 250 ml of a 1.50 M KBr solution?

You know volume and molarity so you plug them into the molarity equation to get moles:

1.5M=mol/.250L*

*Note that I changed 250 ml to .250 L as molarity is determined in liters.

Now we have to switch our equation around to isolate moles:
mol=1.5M x .250L

This gives us an answer of .375 moles. We must now convert moles to grams using the periodic table.
The atomic weight of K (Potassium) is 39.09g/mol, and the atomic weight of Bromine is 79.9g/mol. Adding the two together gives us a molecular weight of KBr, which is 118.99g/mol.

Lastly we must multiply our molecular weight by how many moles we had to get grams of KBr.

118.99g/mol x .375 mol

Moles cancel out giving us a final answer of 44.62g KBr.

Molar Mass Kbr

Moles KBr = 0.250 L x 1.50 M=0.375
molar mass KBr = 119.0 g/mol

mass needed = 119 g/mol x 0.375 mol=44.6 g

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