What happens to the following properties of a pure solvent when a non-volatile s
ID: 993574 • Letter: W
Question
What happens to the following properties of a pure solvent when a non-volatile solute is added? (write "increase" or "decrease" then explain why) i) Vapor pressure will because... ii) Melting point will because... iii) Boiling point will. because... b) If a beaker of pure solvent is placed in a closed container along with another beaker with a solution of a nonvolatile solute dissolved in that solvent: i) What happens to the volumes of each liquid (Increases/decreases/No change)? Pure solvent Solution ii) Explain in terms of vapor pressures why the two liquids behave this way Answer the following questions regarding a solution of carbon dioxide (CO_2) in water (H_2O) a) Do you expect the change in enthalpy of the solution (DeltaH_soin) to be small or large & positive? (circle one of the two answers) b) Explain why this is so.Explanation / Answer
2.a.i) Vapor pressure will decrease because nonvolatile solute reduces the capacity of the solvent molecules to move from the liquid surface to the vapor phase as surface area of the solvent decreases.
ii) Melting point will decrease, because at the melting point vapor pressure of the solid and liquid phase must be the same. The lowering of the vapor pressure leads to the lowering of the temperature at which the vapor pressures of the liquid and frozen forms of the solution will be equal.
iii) Boiling point will increase, because the boiling point is the temperature where the vapor pressure of the solvent becomes equal to the external atmospheric pressure. When solute is added in the solvent then solvent molecules take more time and temperature (as nonvolatile solute reduces the capacity of the solvent molecules to move from the liquid surface to the vapor phase ) to make vapor pressure and causes the boiling point of the solvent increases.
b)i) Pure solvent : decreases.
solution: increases
ii) As dissolved solute decreases the vapor pressure over the solution, so when pure solvent and
solution are kept in a closed container, vapor pressure over the solution will be less than the vapor pressure over the solvent. The vapor pressure necessary to achieve equilibrium with the pure solvent is higher than that required with the solution. The solvent will try to make more vapor over it to gain the equilibrium with the vapor phase while the solution will try to remove vapor molecules to gain the equilibrium. As a net result solvent molecules will transfer from the pure solvent to the solution.
3. a. small (not large and positive)
b. free energy of the process, dG = dH - TdS
as gas molecules go to liquid state to form the solution, dS is highly negative. So the second part of the above equation ( - TdS) becomes positive. To make the process a feasible one (dG < 0 or very near to zero) dH should be small (even negative, for CO2 enthalpy of dissolution is -19.4 kJ/mol). It cannot be very large as then value of dG will very large and positive which will make the reaction non- spontaneous.
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