Q : sample 0.2795 g containing only C(adsbygoogle = window.adsbygoogle || []).push({}); _{6}H_{6}Cl_{6}(Mw = 209.8) and C_{14}H_{9}Cl_{5}(Mw = 354.5)

precipitate AgCl 0.7161 g (Calculate %(C_{6}H_{6}Cl_{6}) and %(C_{14}H_{9}Cl_{5}) in sample)

AgCl has 143.3 g/mol and 4.997 x 10^{-3}

I get

0.2795 g = weight of (C_{6}H_{6}Cl_{6}) + weight of (C_{14}H_{9}Cl_{5})

mol of AgCl = mol of (C_{6}H_{6}Cl_{6}) + mol of (C_{14}H_{9}Cl_{5})

where weight of (C_{6}H_{6}Cl_{6}) = x

4.997 x 10^{-3}= x/(Mw of C_{6}H_{6}Cl_{6}) + (0.2795 - x)/(Mw of_{14}H_{9}Cl_{5})

I have calculated and my answer incorrectly

(The correct answer is 26.01 % (C_{6}H_{6}Cl_{6}) and 77.99% (_{14}H_{9}Cl_{5}))

please help me !!!

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