Question about MPS Multimedia QuickStudy Chemistry for PC
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As an organic compound, lidocaine is insoluble in water, so I am assuming that either some combination of water and alcohol (ethanol), or a pure organic solvent like ethanol or diethyl ether is being used.
If lidocaine were soluble in water (which it's not), you would have been able to solve your problem as indicated below. Maybe you would be able to modify the indicated solution by taking into consideration the density of the organic solvent.
Percentage (%) by mass is one of the common ways to express the concentrations of solutions.
Since % means the number of grams of solute in 100 grams of solution (solute + solvent), a 1.5% solution of any solute refers to 1.5 grams of it in 100 grams of solution.
Assuming we are referring to an aqueous solution (that is, where water is the solvent), this reasonably approximates to 1.5 grams of lidocaine per 100 mL solution. We can assume this result, because we have a dilute watery solution, which probably has the density close to the density of pure water, which is 1.0 gram per 1.0 mL at commonly used temperatures, such as average room temperature.
So, 1.5 g lidocaine per 100 mL solution can easily be scaled down by one-tenth to give:
0.15 g lidocaine per 10 mL solution.
Posted on Jan 04, 2011
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