User:Mary Mendoza/Notebook/CHEM572 Exp. Biological Chemistry II/2013/01/23

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Revision as of 16:27, 23 January 2013

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It was decided to prepare a 30 mM stock solution of adenosine. The calculation below shows the amount needed to obtain 30 mM of adenosine.

.030 $\frac{mol}{L}$ mol of adenosine × $\frac{267.24 g}{1 mol}$ = $\frac{8.0172 g}{L}$ × .010 L = 0.0802 g of adenosine

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	.030 <math>\frac{mol}{L}</math> mol of adenosine × <math>\frac{267.24 g}{1 mol}</math> = <math>\frac{8.0172 g}{L}</math> × .010 L = 0.0802 g of adenosine		.030 <math>\frac{mol}{L}</math> mol of adenosine × <math>\frac{267.24 g}{1 mol}</math> = <math>\frac{8.0172 g}{L}</math> × .010 L = 0.0802 g of adenosine
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