Calculate the mole fraction of benzene in solution containing 30% by mass in ccl4

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. from Chemistry Solutions Class 12 Haryana Board
Calculate the mole fraction of benzene in solution containing 30%

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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 fraction of benzene in solution containing 30% by mass in carbon tetrachloride. from Chemistry Solutions Class 12 Haryana Board

solution; Molarity (M) is defined as number of moles of solutedissolved in one litre (or one cubic decimetre) of solution. (a) Mol. mass of Co ( NO 3 ) . 6 H 2 O = 58 . 9 + ( 14 + 3 × 16 ) 2 + 6 ( 18 ) = 58 . 9 + ( 14 + 48 ) × 2 + 108 = 58 . 9 + 124 + 108 = 290 . 9 Moles of Co ( NO ) 3 . 6 H 2 O = 30 290 . 9 = 0 . 103 mol . Volume of solution = 4.3 L Molarity, M = Moles of solute Volume of solution in litre = 103 4 . 3 = 0 . 024 M (b) Number of moles present in 1000 ml of 0.5M H 2SO 4= 0.5 mol therefore number of moles present in 30ml of 0.5MH 2SO 4= 0 . 5 × 30 1000mol =0.015mol therefore molarity =0.015/0.5L thus molarity is 0.03M