Liquid Solution & Colligative Properties flash cards

1. 1.What is a solution? Name its two components.

   A homogeneous mixture of two or more substances. Components: solvent (larger amount, determines physical state) and solute (smaller amount, dissolved in solvent).

   Hint: Homogeneous, single phase.

2. 2.List the possible types of binary solutions based on the physical state of solute and solvent.

   Nine types: solute (gas/liquid/solid) $\times$ solvent (gas/liquid/solid). Examples: gas in gas (air), gas in liquid ($\text{O}_2$ in water), solid in liquid (sugar in water), solid in solid (alloys), etc.

   Hint: 3 states of solute $\times$ 3 states of solvent.

3. 3.Define molarity $(M)$ and give its units.

   Molarity $= \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$. Units: $\text{mol L}^{-1}$. It is temperature dependent (volume changes with $T$).

   Hint: Per litre of solution, not solvent.

4. 4.Define molality $(m)$ and give its units.

   Molality $= \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$. Units: $\text{mol kg}^{-1}$. It is temperature independent (mass does not change with $T$).

   Hint: Per kg of solvent; T-independent.

5. 5.Define mole fraction of a component in a binary solution.

   For component A: $x_A = \dfrac{n_A}{n_A + n_B}$. Also $x_A + x_B = 1$. It is dimensionless and temperature independent.

   Hint: Ratio of moles; all fractions sum to 1.

6. 6.Define ppm (parts per million) for a dilute solution.

   $\text{ppm} = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$. Used for very dilute solutions (e.g. trace pollutants, hardness of water).

   Hint: Multiply mass ratio by $10^6$.

7. 7.Define normality $(N)$ and relate it to molarity.

   Normality $= \dfrac{\text{gram equivalents of solute}}{\text{volume of solution in L}}$. Relation: $N = M \times n$, where $n$ = n-factor (valency/basicity/acidity/electrons transferred).

   Hint: $N = M \times$ n-factor.

8. 8.Which concentration terms are temperature independent and why?

   Molality, mole fraction, and mass percent — because they depend only on mass/moles, which do not change with temperature. Molarity and normality depend on volume, so they are temperature dependent.

   Hint: Mass-based = T-independent; volume-based = T-dependent.

9. 9.Define mass percent $(\%\,w/w)$ of a solution.

   $\%\,w/w = \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$. Example: 10% glucose means 10 g glucose in 100 g solution.

   Hint: Mass solute per 100 g solution.

10. 10.Relate molarity $(M)$ and molality $(m)$ using density $d$ (g/mL) and molar mass $M_B$ of solute.

    $m = \dfrac{1000\,M}{1000\,d - M\,M_B}$. For dilute solutions where $M M_B \ll 1000d$, $m \approx M/d$.

    Hint: Density bridges volume-based and mass-based.

11. 11.State Henry's law.

    At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid: $p = K_H \, x$, where $x$ = mole fraction of gas in solution, $K_H$ = Henry's constant.

    Hint: $p = K_H x$; solubility $\propto$ pressure.

12. 12.In Henry's law $p = K_H x$, how does a higher $K_H$ affect solubility?

    Higher $K_H$ means lower solubility of the gas (for a given partial pressure, $x = p/K_H$ is smaller). Different gases have different $K_H$ values at the same temperature.

    Hint: $x = p/K_H$: bigger $K_H$, smaller $x$.

13. 13.How does the solubility of a gas in liquid vary with temperature?

    Solubility decreases with increasing temperature (dissolution of gas is exothermic). Hence $K_H$ increases with $T$. This is why aquatic life suffers in warm water (less dissolved $\text{O}_2$).

    Hint: Warm soda goes flat; gas dissolution is exothermic.

14. 14.Give two practical applications of Henry's law.

    (1) Soft drinks/soda are bottled under high $\text{CO}_2$ pressure to increase gas solubility. (2) Scuba divers use diluted air (helium + $\text{O}_2$) to avoid bends caused by dissolved $\text{N}_2$ at high pressure.

    Hint: Soda bottling and deep-sea diving.

15. 15.What is the bends (decompression sickness) and how does Henry's law explain it?

    At depth, high pressure increases $\text{N}_2$ solubility in blood. On rapid ascent, pressure drops, and dissolved $\text{N}_2$ escapes as bubbles in tissues/blood, causing painful bends. Prevented by using helium-oxygen mixtures.

    Hint: Nitrogen bubbles out on fast ascent.

16. 16.State Raoult's law for a solution of two volatile liquids.

    The partial vapour pressure of each component equals its mole fraction times the vapour pressure of the pure component: $p_A = p_A^{\circ} x_A$ and $p_B = p_B^{\circ} x_B$. Total: $p = p_A + p_B$.

    Hint: $p_i = p_i^{\circ} x_i$; partial pressures add.

17. 17.Write the total vapour pressure of a binary solution of volatile liquids A and B in terms of $x_A$.

    $p = p_A^{\circ} x_A + p_B^{\circ} x_B = p_B^{\circ} + (p_A^{\circ} - p_B^{\circ})x_A$. A linear function of $x_A$ between $p_B^{\circ}$ (at $x_A=0$) and $p_A^{\circ}$ (at $x_A=1$).

    Hint: Linear in $x_A$; straight-line plot.

18. 18.State Raoult's law for a solution containing a non-volatile solute.

    The vapour pressure of the solution equals mole fraction of solvent times pure solvent vapour pressure: $p_{soln} = x_{solvent}\, p^{\circ}$. Only the solvent contributes to vapour pressure.

    Hint: Only solvent evaporates; $p = x_{solv}p^{\circ}$.

19. 19.How is Raoult's law a special (limiting) case of Henry's law?

    For the volatile component in the solvent-rich region, both give $p_i \propto x_i$. When $K_H = p_i^{\circ}$, Henry's law $p_i = K_H x_i$ becomes Raoult's law $p_i = p_i^{\circ} x_i$.

    Hint: Set $K_H = p^{\circ}$.

20. 20.Define an ideal solution. Give its three key properties.

    A solution obeying Raoult's law over the entire composition range. Properties: $\Delta_{mix}H = 0$, $\Delta_{mix}V = 0$, and solute–solvent interactions equal solute–solute and solvent–solvent (A–B $\approx$ A–A $\approx$ B–B).

    Hint: $\Delta H = 0$, $\Delta V = 0$, similar interactions.

21. 21.Give two examples of nearly ideal solutions.

    Benzene + toluene; n-hexane + n-heptane; chlorobenzene + bromobenzene. These have very similar molecular structures and intermolecular forces.

    Hint: Structurally similar liquids.

22. 22.What characterizes a solution showing positive deviation from Raoult's law?

    Observed vapour pressure is higher than predicted. A–B interactions are weaker than A–A and B–B. $\Delta_{mix}H > 0$ (endothermic), $\Delta_{mix}V > 0$ (expansion).

    Hint: Weaker new forces $\Rightarrow$ escape more easily, higher VP.

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