Calculate the mole fraction of benzene in solution containing 30 by mass in carbon tetrachloride

  1. Calculate the mole fraction of benzene in a solution containing 30 % by mass of it in carbon tetrachloride.
  2. Calculate the mole fraction of benzene in solution containing 30% by m


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Calculate the mole fraction of benzene in a solution containing 30 % by mass of it in carbon tetrachloride.

Let us start with 100 of the solution in which Mass of benzene = 30 g Mass of carbon tetrachloride = 70 g Molar mas of benzene `(C_(6)H_(6))=6xx12+6xx1=78" g mol"^(-1)` `n_(C_(6)H_(6))=((30g))/((78 " g mol"^(-1)))=0.385` mol Molar mass of carbon tetrachloride `(C Cl_(4))=12+4 xx 35.5 = 154 "g mol"^(-1)` `n_( C Cl_(4))=((70g))/((154"g mol"^(-1)))=0.454` mol `x_(C_(6)H_(6))=(n_(C_(6)H_(6)))/(n_(C_(6)H_(6))+n_(C Cl_(4)))=((0.385" mol"))/((0.385" mol")+(0.454" mol"))=0.459`.

Calculate the mole fraction of benzene in solution containing 30% by m

W C 6 H 6 = 30 g , W s o l = 100 g , W C C l 4 = ( 100 − 30 ) g = 70 g M w of C 6 H 6 = 78 g m o l − 1 , M w o f C C l 4 = 12 + 4 × 35.5 = 154 g m o l − 1 n C 6 H 6 = W C 6 H 6 M w o f c 6 H 6 = 30 g 78 g m o l − 1 = 0.385 n C C l 4 = W C C l 4 M w o f C C l 4 = 70 g 154 g m o l − 1 = 0.425 C H M i C 6 H 6 = n C 6 H 6 n C 6 H 6 + n C C l 4 = 0.385 0.385 + 0.425 = 0.385 0.84 = 0.485 C H M i C C l 4 = 1 − 0.485 = 0.542

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