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

  1. Calculate the mole fraction of benzene is solution containing `30%` by mass in carbon tetrachloride. Then calculate the mole fraction by using the for
  2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
  3. calculate the mole fracton of benzene ina solution containing 30 % by mass of it in carbon tetrachloride


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Calculate the mole fraction of benzene is solution containing `30%` by mass in carbon tetrachloride. Then calculate the mole fraction by using the for

For 100 g of the solution Mass of benzene = 30 g Mass of carbon tetrachloride `=100-30=70"g"` Molar mass of benzene `(C_(6)H_(6))=(12xx6)+(6xx1)=72+6=78"g mol"^(-1)` Moles of benzene, `n_(C_(6)H_(6))=("Mass")/("Molar mass")=(30)/(78)=0.385" mol"` Molar mass of carbon tetrachloride `(C Cl_(4))=12+(35.5xx4)` `=12+142.0=154" g mol"^(-1)` Moles of `C Cl_(4),n_(C Cl_(4))=(70"g")/((154" mol"^(-1)))=0.454" mol"` Mole fraction of benzene, `x_(C_(6)H_(6))=(n_(C_(6)H_(6)))/(n_(C_(6)H_(6))+n_(C Cl_(4)))=(0.385" mol")/((0.385+0.454)" mol")=0.459`

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

Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴Mass of carbon tetrachloride = (100 − 30)g = 70 g Molar mass of benzene (C 6H 6) = (6 × 12 + 6 × 1) g mol −1 = 78 g mol −1 ∴Number of moles of `C_6H_6 =30/78 mol` = 0.3846 mol Molar mass of carbon tetrachloride (CCl 4) = 1 × 12 + 4 × 355 = 154 g mol −1 ∴Number of moles of CCl 4 = 70/154 mol = 0.4545 mol Thus, the mole fraction of C 6H 6is given as: `("Number of moles of " C_6H_6)/("Number of moles of " C_6H_6+"Number of moles of" C Cl_4)` `= 0.3846/(0.3846+0.4545)` = 0.458

calculate the mole fracton of benzene ina solution containing 30 % by mass of it in carbon tetrachloride

Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴ Mass of carbon tetrachloride = (100 - 30)g = 70 g Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol - 1 = 78 g mol - 1 ∴ Number of moles of C6H6 =30/78 mol = 0.3846 mol Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 355 = 154 g mol - 1 ∴ Number of moles of CCl4 = 70/154 mol = 0.4545 mol Thus, the mole fraction of C6H6 is given as: = 0.3846 / (0.3846 +0.4545) = 0.458

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