Chemistry
posted by Tawnya .
Can someone please help me out here?
I've tried several times and i keep getting the wrong answer :( lol
A small crucible is filled with 100.00 mL of nhexadecane (MW (226.44 g/mol), density (0.7703 g/mL))
and 2.10 L of 7octenyltrichlorosilane (MW (245.65 g/mol), density (1.07 g/mL)).
Calculate the concentration in mM of the reagent in solution

Number of moles of nhexadane is
100 mL * (0.7703 g/mL) * (1 mol / 226.44 g)
mL and g units cancel
= 100 * (0.7703/226.44) moles nhexadane
The total volume of the final solution is 100 mL + 2100 mL = 2200 mL = 2.2 L
The concentration of nhexadane in mM is
100 * (0.7703/226.44) M / 2.2L * (1000 mM / M)
If nhexadane is your reagent, then this is the final answer; otherwise repeat this type of calculation to find the final concentration of 7ocenyltrichlorosilane in solution.
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