Question

You are preparing a solution of lithium bromide to use during an experiment. The experiment calls for 909.7 mL a 18 M solution. How many moles of the solute must be used to achieve this?

Answer 1

16.37 moles of lithium bromide.

Step-by-step explanation:

909.7 mL of 18 M

= .9097 L of 18 M solution of lithium bromide .

no of moles required = .9097 x 18

= 16.37 moles of lithium bromide.