An organic compound is subjected to chlorinat | vedclass

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Question

(A) The molar mass of $AgCl = 107.87 + 35.5 = 143.37 \ g/mol$.The number of moles of $AgCl$ formed $= \frac{0.3849 \ g}{143.37 \ g/mol} = 0.0026846 \

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(A) The molar mass of $AgCl = 107.87 + 35.5 = 143.37 \ g/mol$.The number of moles of $AgCl$ formed $= \frac{0.3849 \ g}{143.37 \ g/mol} = 0.0026846 \ mol$.Since $1 \ mol$ of $AgCl$ contains $1 \ mol$ of $Cl$,the moles of $Cl = 0.0026846 \ mol$.The mass of chlorine $= 0.0026846 \ mol 	imes 35.5 \ g/mol = 0.0953 \ g$.The percentage of chlorine in compound $A = \frac{	ext{mass of chlorine}}{	ext{mass of compound } A} 	imes 100$.$\% \ Cl = \frac{0.0953 \ g}{0.5 \ g} 	imes 100 = 19.06 \ \%$.Rounding off to the nearest integer,we get $19$.

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