Tuesday

September 16, 2014

September 16, 2014

Posted by **Tawnya** on Monday, January 21, 2013 at 1:03am.

I've tried several times and i keep getting the wrong answer :( lol

A small crucible is filled with 100.00 mL of n-hexadecane (MW (226.44 g/mol), density (0.7703 g/mL))

and 2.10 L of 7-octenyltrichlorosilane (MW (245.65 g/mol), density (1.07 g/mL)).

Calculate the concentration in mM of the reagent in solution

- Chemistry -
**anonymous**, Tuesday, January 22, 2013 at 8:27pmNumber of moles of n-hexadane is

100 mL * (0.7703 g/mL) * (1 mol / 226.44 g)

mL and g units cancel

= 100 * (0.7703/226.44) moles n-hexadane

The total volume of the final solution is 100 mL + 2100 mL = 2200 mL = 2.2 L

The concentration of n-hexadane in mM is

100 * (0.7703/226.44) M / 2.2L * (1000 mM / M)

If n-hexadane is your reagent, then this is the final answer; otherwise repeat this type of calculation to find the final concentration of 7-ocenyltrichlorosilane in solution.

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