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Old 22-07-2009, 05:55 PM   #14
shadow-light
He was no dragon. Fire cannot kill a dragon
 
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Join Date: Jul 2007
Location: York
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ok... I think I've found the right set of lecture notes... I've found these 2 examples so far of similar types of calculation... my maths fails, but I thought I'd put them up in case they help anyone work anythi gout...

I'll keep looking for the exact manor for doing this


Quote:
General formula for solving: Amt1 x %1 = Amt2 x %2
Amt1 = the quantity or amount of the ORIGINAL preparation.
%1 = the % strength of the ORIGINAL preparation expressed as a decimal or percent
Amt2 = the quantity or amount of the WANTED preparation
%2 = the % strength of the WANTED preparation expressed as a decimal or percent
To solve concentration and dilution problems you need to identify the two preparations in the equation, convert ratio or percentage strengths to decimal expressions and convert to same systems of measurement.

EXAMPLE: If 500 ml of a 15% solution are diluted to 1500 ml, what will be the percent strength?

Amt1 x %1 = Amt2 x %2
Step 1: Identify the two preparations in the problem and assign values to appropriate terms.
Step 2: Solve the equation by multiplying and solving for X.
500 ml x 15% = 1500 ml x X%
7500 = 1500X
7500/1500 = 1500X/1500
X = 5%
Quote:
If 8 ml of phenol were added to 480 ml of lotion, what is the percentage of phenol in the lotion?
8 ml (Active Ingredient)

(Total Amt) 480 ml X %
Divide to solve for X.
8 / 480 = X

EXAMPLE: If 1.2 gm of menthol is added to 480 ml of lotion, what is the percentage of menthol in the lotion?

1.2 gm (Active Ingredient)
(Total Amt) 480 ml X %
Divide to solve for X.
1.2 / 480 = x
x = 0.0025

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