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Topic: 4% (Poly(L-Lactide)) solution with Dichloromethane  (Read 1988 times)

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Offline vanillafinger83

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4% (Poly(L-Lactide)) solution with Dichloromethane
« on: January 09, 2012, 12:13:49 PM »
Is my calculation below correct to make 10mL of 4% (Poly(L-Lactide)) in dichloromethane solution?

1. total dilution volume: 10mL
2. density of dichloromethane: 1.33g/cm3
3. weight of dichloromethane solution needed: 10mL x 1.33g/cm3 = 13.3g
4. weight of PLLA powder needed to make 4% PLLA solution: 13.3g x (4/100) = 0.532g

Then mix 0.532g PLLA (s) into 10mL of dichloromethane (l) to make PLLA - dichloromethane with PLLA concentration of 4% (w/v)?

Can someone check this work for me?  Thanks!!


Offline fledarmus

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Re: 4% (Poly(L-Lactide)) solution with Dichloromethane
« Reply #1 on: January 09, 2012, 12:41:18 PM »
As I understand it, the calculations you have worked through give you a % w/w solution rather than a % w/v solution. Weight/volume percent solutions are odd in that they actually have units, usually g/mL. All you need for a 4% w/v solution is 0.04 g/mL. This is read "0.04 grams of solute per milliliter of solution" (not "per milliliter of solvent"). The mass of the solvent doesn't come into consideration, only the total volume of the solution.

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