Chemistry Solutions Molarity
Now the #"moles of solute"# are a constant. The volume of solution MAY change substantially with increasing or decreasing temperature. In some calculations #"molality"# is used in preference, which is defined by the quotient.... #"molality"="moles of solute"/"kilograms of solvent"# ....this is temperature independent, and at lower concentrations, #"molarity"-="molality"#.
To get the molarity, you divide the moles of solute by the litres of solution.
#"Molarity" = "moles of solute"/"litres of solution"#
For example, a 0.25 mol/L NaOH solution contains 0.25 mol of sodium hydroxide in every litre of solution.
To calculate the molarity of a solution, you need to know the number of moles of solute and the total volume of the solution.
To calculate molarity:
- Calculate the number of moles of solute present.
- Calculate the number of litres of solution present.
- Divide the number of moles of solute by the number of litres of solution.
EXAMPLE:
What is the molarity of a solution prepared by dissolving 15.0 g of NaOH in enough water to make a total of 225 mL of solution?
Solution:
1 mol of NaOH has a mass of 40.00 g, so
#"Moles of NaOH" = 15.0 cancel("g NaOH") × "1 mol NaOH"/(40.00 cancel("g NaOH")) = "0.375 mol NaOH"#
#"Litres of solution" = 225 cancel("mL soln") × "1 L soln"/(1000 cancel("mL soln")) = "0.225 L soln"#
#"Molarity" = "moles of solute"/"litres of solution" = "0.375 mol"/"0.225 L" = "1.67 mol/L"#
Some students prefer to use a "molarity triangle".
It summarizes the molarity formulas as
#"Moles" = "molarity × litres"#
#"Molarity" = "moles"/"litres"#
#"Litres" = "moles"/"molarity"#