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Q : sample 0.2795 g containing only C6H6Cl6 (Mw = 209.8) and C14H9Cl5 (Mw = 354.5)
precipitate AgCl 0.7161 g (Calculate %(C6H6Cl6) and %(C14H9Cl5) in sample)
AgCl has 143.3 g/mol and 4.997 x 10-3
I get
0.2795 g = weight of (C6H6Cl6) + weight of (C14H9Cl5)
mol of AgCl = mol of (C6H6Cl6) + mol of (C14H9Cl5)
where weight of (C6H6Cl6) = x
4.997 x 10-3 = x/(Mw of C6H6Cl6) + (0.2795 - x)/(Mw of 14H9Cl5)
I have calculated and my answer incorrectly
(The correct answer is 26.01 % (C6H6Cl6) and 77.99% (14H9Cl5))
please help me !
precipitate AgCl 0.7161 g (Calculate %(C6H6Cl6) and %(C14H9Cl5) in sample)
AgCl has 143.3 g/mol and 4.997 x 10-3
I get
0.2795 g = weight of (C6H6Cl6) + weight of (C14H9Cl5)
mol of AgCl = mol of (C6H6Cl6) + mol of (C14H9Cl5)
where weight of (C6H6Cl6) = x
4.997 x 10-3 = x/(Mw of C6H6Cl6) + (0.2795 - x)/(Mw of 14H9Cl5)
I have calculated and my answer incorrectly
(The correct answer is 26.01 % (C6H6Cl6) and 77.99% (14H9Cl5))
please help me !