Molar Concentration of Molarity of Solutions

FIRST OFF, WELCOMEE BACK! THIS IS THE VERY FIRST BLOG OF 2011! I HOPE ALL OF YOU HAVE AN AWESOME BREAK! I KNOW YOU GUYS ALL MISSED THE ATOMIC THREE! WELL TIME TO GET TO WORK!


So last class we discussed material about molar concentration or known as molarity of solutions. Here are some notes that are a summary of last days class.


Molar concentration is concentration measured by the number of moles of solute per liter of solution.


Formula:
Molar concentration (M) =  Moles of solute (n)   /   Volume of solution (V in liter)


This formula can be flipped to find the volume and or the number of moles. 


moles of solute= molarity X volume of solution 
Volume of solution= moles of solute/molarity

Examples:

#1 Calculate the molarity when NaOH has 0.510 moles and the volume is 1.400. Be sure to use the right number of significant figures. 

M= 0.510moles of NaOH/ 1.400L

=0.364 moles/ L

#2
Q: Find out the molar concentration of acetic acid (CH₃COOH) solution when  1.24 grams of acetic acid (CH₃COOH) dissolved in 155.0 ml solution?
Solution:  Given,
mass of acetic acid (w)  =  1.24 g
volume of acetic acid (V) = 155.0 ml
                                      = (155.0 / 1000) L         {Because  1 Liter  = 1000 ml}
                                     = 0.155 L
Molar mass of acetic acid (W)  = atomic mass of C atom  + atomic mass of 2 O atoms  + atomic mass of 4 H atoms
                                                        = (12.0  + 2 * 16.00  + 4 * 1.0) g / mol                                                  
                                                        = 48.0 g / mol
And now by the help of molar concentration (molarity) formula we can determine molar concentration of  acetic acid,


Molar concentration of acetic acid (M)  = mass of acetic acid (w)   /   molar mass of acetic acid (W) *  volume of acetic acid (V)
                                                                      =  1.24 g   /  (48.0 g / mol  *  0.155 L)
                                                                      = 0.167 M
So when  1.24 grams of acetic acid (CH₃COOH) dissolved in 155.0 ml solution, molar concentration of this solution is 0.167 M.

 See you guys next class :)!