Determine the mole fraction of chloroform in a solution produced by adding 1120
ID: 825474 • Letter: D
Question
Determine the mole fraction of chloroform in a solution produced by adding 1120 ml of CHCl3 to 6.18 kg of benzene.
Determine the mole fraction of CH3CO2H in a solution produced by combining 0.0995 kg of acetic acid to 2080 g of H2O.
A solution of benzene in the solvent tetrahydrofuran has a mole fraction of C6H6 of 0.378. What mass (g) of the solution must be taken if1870 g of benzene is needed?
A 9090 g solution of hexane and benzene contains 0.839 L of CH3(CH2)4CH3. Calculate the mole fraction of hexane in the solution.
etermine the mole fraction of C4H8O in a solution produced by adding 694 ml of tetrahydrofuran to 10500 ml of (C2H5)2O.
Determine the mole fraction of acetone in a solution made by adding 0.270 L of (CH3)2CO to 328 g of H2O.
Molar Mass (g/mol) CHCl3 119.38 C6H6 78.12 Density (g/ml): CHCl3 1.483 C6H6 0.8786 Name/Formula: chloroformCHCl3
benzene
C6H6
Explanation / Answer
a) mole fraction of CHCl3 = (1120* 1.483/119.38) / (1120*1.483/119.38) + (6180/ 78.12) = 0.14956
b) mole fraction = (99.5 / 60.05) / (99.5 / 60.05) + (2080 / 18.015) = 0.01415
c) mole of benzene = 1870 / 78.12 = 23.938
mass of tetrahydrofuran = (23.938 / 0.378)* 0.622*72.12 = 2840.8 g
total mass = 4710.8 g
d) mass of hexane = 839 * 0.6838 = 573.708 g
mass of benzene = 9090 - 573.708 = 8516.292 g
mole fraction of hexane = (573.708 / 100.21) / [(573.708 / 100.21) + ( 8516.292 / 78.12)] = 0.05
e) moles of tetrahydrofuran = 0.8894 * 694/ 72.12 = 617.2436/ 72.12 = 8.5586
moles of (C2H5)2O = 0.7138 * 10500 / 74.12 = 7494.9 / 74.12 = 101.1185
mole fraction of tertrahydrofuran = 8.5586 / (8.5586 + 101.1185) = 0.078
f) moles of acetone = 0.27 * 0.7899 / 58.05 = 3.674
moles of water = 328 / 18.015 = 18.207
mole fraction of acetone = 3.674 / (3.674 + 18.207 ) = 0.168
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