Asked by Nomonde langa
                I was asked to prepare a 1.0 x 10-3M Iodine solution in Methylene Chloride.  How do I go by this calculation? (Note: ............10 raised to the power of -3)
            
            
        Answers
                    Answered by
            DrBob222
            
    How much are you to prepare? 1L, 500 mL, 250 mL?
Let's make a liter of the solution. You can adjust from there if you don't want that much.
M = moles/L
1 x 10^-3 M = moles/L
moles = 1 x 10^-3 M x 1 L = 1 x 10^-3
So you want 1 x 10^-3 moles.
Next, moles = grams/molar mass
solve for grams.
g = moles x molar mass
g = 1 x 10^-3 moles x 254 (you look up molar mass I2--I've estimated) = 0.254 grams I2.
Do you know how to make it up. You weigh 0.254 g I2, add the solid to a 1 L volumetric flask, add a little solvent, swirl to dissolve ALL of the solid, then add solvent to the mark of the flask. Please note that this is NOT th same as adding 0.254 g I2 and 1000 mL solvent to the flask.
    
Let's make a liter of the solution. You can adjust from there if you don't want that much.
M = moles/L
1 x 10^-3 M = moles/L
moles = 1 x 10^-3 M x 1 L = 1 x 10^-3
So you want 1 x 10^-3 moles.
Next, moles = grams/molar mass
solve for grams.
g = moles x molar mass
g = 1 x 10^-3 moles x 254 (you look up molar mass I2--I've estimated) = 0.254 grams I2.
Do you know how to make it up. You weigh 0.254 g I2, add the solid to a 1 L volumetric flask, add a little solvent, swirl to dissolve ALL of the solid, then add solvent to the mark of the flask. Please note that this is NOT th same as adding 0.254 g I2 and 1000 mL solvent to the flask.
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