Question

How many grams of KBr are required to make 350 mL of a .115M KBr solution?

Answer 1

use the direct molarity fomula

molarity M =( W / MW ) x 100 / V

where

M = Molarity = 0.115

W = weight of solute (KBr) = ?

MW = Molecular weight of solute (KBr) = 119.0023 g/mol

V = volume of the solution = 350 mL

plug in all the values in the above equation

0.115 M = (W / 119.0023 g / mol ) x 1000 / 350 mL

W = 0.115 x 119.0023 x 350 / 1000

= 4.79 gr