How many grams of CaBr2 would be needed to create 450.0 mL of a 2.00 M solution?

Answer 1

Begin with the equation for molarity.

M = moles of solute/ L solution.

Plug in what we know.

M = 2 M

m = ?

L = 0.450 L

2 M = m/ 0.45 L

Rearrange the equation to solve for m:

m = 2 M x 0.45 L = 0.9 moles CaBr2

The atomic mass of CaBr2 is 199.89 g/ mole. Use dimensional analysis to determine the amount of grams:

0.9 moles CaBr2 x 199.89 g/ mole = 180 g.