Respuesta :

Willchemistry
Molar mass calcium bromide :

CaBr2 = 40 + 80 * 2 = 200 g / mol

Volume = 50.0 mL  :  0.05 L

Molarity = 0.25 M

Mass of solute = ?

M = m / mm * V

0.25 = m / 200 * 0.05

m = 0.25 * 200 * 0.05

m = 2.5 g of CaBr2

hope this helps!.


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