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Sergeeva-Olga [200]
3 years ago
12

A standard solution is prepared for the analysis of fluoxymesterone (c20h29fo3; 336 g/mol), an anabolic steroid. a stock solutio

n is first prepared by dissolving 16.8 mg of fluoxymesterone in enough water to give a total volume of 500.0 ml. a 1.0 μl aliquot (portion) of this stock solution is taken out and then diluted to a final volume of 1.0 ml. calculate the final concentration of this new solution in m.
Chemistry
1 answer:
-Dominant- [34]3 years ago
3 0

<em>Answer:</em>

  • The concentration of new solution will be 1×10∧-7 M.

<em>Solution:</em>

<em>Data Given </em>

       given mass of fluoxymesterone =16.8mg = 0.0168 g

       molar mass  of fluoxymesterone = 336g/mol

       vol. of fluoxymesterone = 500.0 ml = 0.500 L

      Stock Molarity of  fluoxymesterone = (0.0168/336)÷0.500 = 1×10∧-4 M

So applying dilution formula

                   Stock Solution :  New Solution

                                 M1.V1 = M2.V2

       ( 1×10∧-4 M) × (1×10∧-6 L) = M2 × 0.001 L

     [( 1×10∧-4) × (1×10∧-6)]÷[0.001] = M2

     1 × 10∧-7 = M2

<em>Result:</em>

  • The concentration of new solution M2 will be  1 × 10∧-7
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Explanation:

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\text{Moles of Na} =  \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}

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