Assume that paricalcitol is the limiting reagent in the hydrogenation reaction s

Question

Assume that paricalcitol is the limiting reagent in the hydrogenation reaction shown above. What is the percent yield of the product if you start with 5.0 grams of paricalcitol and you obtain 4.0 grams of isolated product? Most organic reactions first involve dissolving a starting compound in an excess of solvent. Suppose you want to prepare 100 ml of a methanol solution that is 0.10 mM in paricalcitol. How many milligrams of paricalcitol should you add to the methanol? a 0.10 mM solution contains 0.10 millimoles of solute per liter of solution

Explanation / Answer

9.

1 mol of paricalcitol = 1 mol of product

416.636 g. of parcalcitol = 422.636 g. of product

then, 5 g. of paricalcitol = 422.636 x 5 / 416.636 = 5.072 g. of product should be obtained.

Therefore % yield = 4.0 x 100 / 5.072 = 78.86 %

10.

We know that, Molarity =

(weight of solute x 1000) / (gram molecular weight of solute x volume of solution in mL)

weight of solute = (M x gmw x V ) / 1000

w = ( 0.0001 x 416.636 x 100 ) / 1000

w = 0.0042 g . = 4.2 mg. of paricalcitol is to be taken.

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