Solving the weight of substances needed to produce molar and normal solutions

To be able to solve the weight of substances needed to prepare certain molar and normal solutions, you can use the general formulas:

For Normal solutions

N = GEW/L of solution

GEW = W/MW/v

N = (W/MW/v)/L of solution

Where:

GEW = Gram Equivalent Weight

L = Liter

W = weight in grams of substance

EW = Equivalent Weight

MW = Molecular Weight

v = valence

The short cut formula is:

W = DN X DV X EW (MW/v)

Where:

DN = Desired Normality

DV = Desired Volume
EW = Equivalent Weight

MW = Molecular weight

v = valence

Here’s a sample problem.

How much CaCl2 will you need in preparing 500 mL of a 0.5 N solution?

W = DN X DV X EW (MW/v)

W = 0.5 X 0.5 X (111/2)

W = 13.875 grams of CaCl2

To prepare the 0.5 N CaCl2 solution:

Weigh 13.875 grams of CaCl2 and dilute it to 500 mL of solution in  a volumetric flask. You can first dispense 250 ml of the distilled water to the flask, dissolve the 13.875 CaCl2, and then add the diluent up to the 500 mL mark of the volumetric flask.

Take note of the following:

1.    Volume must always be converted to liters when using this formula, or if you don’t want to convert, divide your answer by 1,000.

2.    The powder must not be added to 500 ml or 0.5 L but diluted TO 500 mL to in a volumetric flask to get the exact volume. The resulting volume in your 500 ml flask after dissolving the CacL2 must not be more or less than 500 ml. This will ensure accuracy of your measured solution.

For Molar solutions

M = GMW/L of solution

GMW = Weight/MW

M = (W/MW)/L of solution

Where:

GMW = Gram Molecular Weight

L = Liter

W = weight in grams of substance

MW = Molecular Weight

v = valence


Hence for Molar solutions the formula is:

W = DM X DV X MW

Where:

DM = Desired Molarity
DV = Desired Volume
MW= Molecular Weight

If you're given the same data but asked to solve for the molarity this is the formula and substitution:

Here’s a sample problem.

How much CaCl2 will you need in preparing 500 mL of a 0.5 M solution?

W = DN X DV X MW

W = 0.5 X 0.5 X 111

W = 27.75 grams of CaCl2

 
The only difference from the Normal solution is the absence of valence. 


To prepare the 0.5 M CaCl2 solution:

Weigh 27.75 grams of CaCl2 and dilute it to 500 mL of solution in  a volumetric flask. 

You can first dispense 250 ml of the distilled water in the flask, dissolve the 27.75 CaCl2, and then add the diluent up to the 500 mL mark of the volumetric flask.