Question

How many mL of a 0.160 M aqueous solution of copper(II) bromide, CuBr2, must be taken to obtain 5.86 grams of the salt? __ mL

Answer 1

Molarity = number of moles of solute (1000) / volume of solvent in ml

Given, salt = 5.86 g ; molarity = 0.160 M

      number of moles of solute = given mass /molar mass

Molar mass of Cu = 63.546 g and Br = 79.904 g

Molar mass of Copper(II) bromide; 63.546 +2(79.904) = 223.534 g/mol

      Number of moles               = 5.86g / 223.354 gmol^-1

                                             = 0.0262

                      molarity = 0.0262 (1000) / vol. in ml

                     0.160 M = 0.0262 (1000) / vol. in ml

                  vol. in ml = moles (1000) / molarity

                 vol. in ml = 0.0262 (1000) / 0.160

   vol. in ml = 26.2 /0.160

                vol. in ml =163.75 ml