1. lodine, , is more soluble in dichloromethane, CH CE, than in water because, A
ID: 538545 • Letter: 1
Question
1. lodine, , is more soluble in dichloromethane, CH CE, than in water because, A Both iodine and dichloromethane have strong ion-dipole interactions. B) The dipole-dipole forces in dichloromethane are much stronger than the dispersion forces in iodine C) The intermolecular forces are similar in both iodine and dichloromethane. D) lodine is polar and dichloromethane has a large number of hydrogen bonds 2. KBr does not dissolve well in nonpolar solvents because A) solute-solute interactions are much larger than solvent-solvent or solute-solvent B) solvent-solvent interactions are much larger than solute-solvent or solute-solute interactions C) solute-solvent interactions are much larger than solvent-solvent or solute-solute D) solute-solvent interactions are similar to solvent-solvent and solute-solute interactions. 3. A solution is prepared by dissolving 17.75g of H:SOs in enough water to make 100.0 ml of solution. If the density of the solution is 1.1094 giml, what is the mole fraction H:S0 in the solution? A)0.0181 B) 0.0338 C) 0.0350 D) 19.0 4. To make a 2.00 m solution, one could take 2.00 moles of solute and add A) 100L of solvent B) 1.00 kg of solvent C) Enough solvent to make 1.00 L of solution D) Enough solvent to make 1.00 kg of solutionExplanation / Answer
1) Iodine occurs in Group 17, period 5 of the periodic table. As such, iodine is largely non-polar due to its large size and reduced nuclear attraction for the outer electrons. Dichlormethane is non-polar, since the individual bond moments cancel each other, giving a net dipole moment of zero. Therefore, both iodine and dichloromethane are non-polar and have similar intermolecular forces (mostly Vander Waals’ forces and/or dispersion forces). Thus, iodine is more soluble in dicloromethane and less in water, since water is a polar solvent. Hence, (C) is the correct answer.
2) KBr is a polar molecule. The important intermolecular forces operating in KBr are ionic attractions or ion-ion attractions. The predominant intermolecular forces in non-polar solvents are Vander Waal’s attractions and/or dispersion forces. Therefore, there are two types of interactions in a solution of KBr in a non-polar solvent. The solute-solute interactions between K+ and Br- ions (in KBr) are far greater than the interactions between K+-solvent or Br--solvent. Therefore, (A) is the correct answer.
3) Volume of the solution is 100.0 mL; density of the solution is 1.1094 g/mL.
Mass of the solution = (100.0 mL)*(1.1094 g/mL) = 110.94 g.
Mass of solute, H2SO4 = 17.75 g.
Mass of solvent (water) = (110.94 – 17.75) g = 93.19 g.
Molar mass of H2SO4 = (2*1.008 + 1*32.065 + 4*15.9994) g/mol = 98.0786 g/mol.
Moles of H2SO4 = (mass of H2SO4)/(molar mass of H2SO4) = (17.75 g)/(98.0786 g/mol) = 0.18098 mole.
Molar mass of water, H2O = (2*1.008 + 1*15.9994) g/mol = 18.0154 g/mol.
Moles of H2O = (mass of H2O)/(molar mass of H2O) = (93.19 g)/(18.0154 g/mol) = 5.17279 mole.
Mole fraction of H2SO4 = (moles of H2SO4)/(moles of H2SO4 + moles of H2O) = (0.18098 mole)/(0.18098 mole + 5.17279 mole) = 0.0338
Ans: (B)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.