Questions: Solutions

Q1 Distinguish between molarity and molality. Which one is used for colligative properties and why?

Answer (English): Molarity counts moles of solute against one litre of solution. Molality counts the same moles against one kilogram of the solvent. Colligative properties are calculated with molality because their size depends on the amount of solvent present and on the number of solute particles, not on the volume of the solution. The volume of a solution changes when it is heated or cooled, so a molarity changes with temperature and would make the property temperature dependent twice over. Molality, built from masses, stays the same at any temperature. That is why the formulas are written with m.

Hinglish: Molarity moles per litre solution hoti hai aur temperature ke saath badalti hai, kyunki volume change hota hai. Molality moles per kg solvent hoti hai, mass temperature se nahi badalta, isliye fixed rehti hai. Colligative properties particles ki ginti aur solvent ki mass par depend karti hain, isliye wahan molality use karte hain.

Q2 A solution is labelled 10 percent w/w glucose. Calculate its molality, the mole fraction of glucose and its molarity if the density is 1.2 g mL^-1.

Answer (English): Take 100 g of solution. It has 10 g of glucose and 90 g of water. Glucose is 10 divided by 180, or 0.0556 mol, and water is 90 divided by 18, or 5.000 mol. Molality is 0.0556 divided by 0.0900 kg, which is 0.617 mol kg^-1. The mole fraction of glucose is 0.0556 divided by 5.0556, which is 0.0110, and the mole fraction of water is 0.9890. The volume of 100 g of solution is 100 divided by 1.2, or 83.3 mL, so molarity is 0.0556 divided by 0.0833 L, which is 0.667 M.

Hinglish: 100 g solution lo: 10 g glucose aur 90 g paani. Glucose ke 0.0556 mol aur paani ke 5 mol bante hain. Molality 0.0556/0.0900 = 0.617 m, mole fraction 0.0556/5.0556 = 0.0110, aur volume 100/1.2 = 83.3 mL se molarity 0.667 M. Pehle sample fix karna hi aadha solution hai.

Q3 State Henry's law. Why do scuba divers breathe air diluted with helium, and why does warm water hold less dissolved oxygen?

Answer (English): Henry's law fixes the temperature and ties the partial pressure of the gas over a solution to the gas mole fraction inside the solution: p equals K_H times x. At depth a diver breathes gas at high pressure, so a large amount of nitrogen dissolves in the blood. Nitrogen is only sparingly soluble, but on a fast ascent the excess escapes as bubbles and causes decompression sickness. Helium has a much larger Henry constant and is far less soluble, so replacing part of the nitrogen with helium keeps the dissolved gas low. Dissolving a gas is exothermic, so raising the temperature shifts the equilibrium towards the gas phase and lowers the dissolved oxygen in warm water, which is why thermal pollution harms fish.

Hinglish: Henry's law: p = K_H x, matlab fixed temperature par gas ka partial pressure aur solution mein uski mole fraction proportional hote hain. Gehre paani mein pressure zyada hota hai isliye nitrogen blood mein zyada dissolve ho jaati hai; helium ki solubility kam hoti hai isliye wo safer hai. Gas dissolve karna exothermic hai, isliye garam paani mein oxygen kam rehti hai.

Q4 Derive the expression for the relative lowering of vapour pressure in the presence of a non-volatile solute.

Answer (English): Let the solvent have mole fraction x1 and vapour pressure p1^0 when pure. By Raoult's law the partial pressure of the solvent over the solution is p1 equal to x1 p1^0. For a binary solution x1 plus x2 equals 1, where x2 is the mole fraction of the non-volatile solute. Subtracting p1 from p1^0 gives p1^0 minus x1 p1^0, which is p1^0 times x2. Dividing both sides by p1^0 gives (p1^0 minus p1) divided by p1^0 equal to x2. The relative lowering of vapour pressure therefore equals the mole fraction of the solute, it does not depend on the identity of the solute, and it is a colligative property.

Hinglish: Raoult's law se p1 = x1 p1^0. Binary solution mein x1 + x2 = 1 hota hai. Isliye (p1^0 - p1)/p1^0 = x2 nikalta hai. Yeh result solute pe depend nahi karta, sirf uske mole fraction par depend karta hai, isliye yeh colligative property hai.

Q5 Differentiate ideal and non-ideal solutions. Give one example each of positive and negative deviation.

Answer (English): An ideal solution obeys Raoult's law over the whole composition range, its enthalpy of mixing is zero, its volume of mixing is zero, and the forces between unlike molecules are about equal to those between like molecules; benzene and toluene come close. A non-ideal solution deviates. When the unlike forces are weaker, more molecules escape, the vapour pressure is higher than predicted, and the mixture absorbs heat on mixing; this is a positive deviation, as with ethanol and acetone. When the unlike forces are stronger, the vapour pressure is lower than predicted and heat is released on mixing; this is a negative deviation, as with chloroform and acetone, where a hydrogen bond forms between the chloroform hydrogen and the acetone oxygen.

Hinglish: Ideal solution Raoult's law ko poore range mein follow karta hai, mixing par enthalpy aur volume dono zero, jaise benzene-toluene. Positive deviation mein unlike forces kamzor hote hain, vapour pressure zyada aur mixing endothermic; ethanol-acetone example. Negative deviation mein unlike forces strong, vapour pressure kam aur mixing exothermic; chloroform-acetone H-bonding ki wajah se.

Q6 What is an azeotrope? Why can a mixture of ethanol and water not be separated completely by fractional distillation?

Answer (English): An azeotrope is a mixture whose vapour has the same composition as its liquid, so it boils at a constant temperature and distils without a change in composition. Ethanol and water show a large positive deviation and form a minimum boiling azeotrope at about 95 percent ethanol that boils below either pure liquid. In fractional distillation the vapour is always richer in the more volatile component only until the azeotropic composition is reached. After that the distillate no longer changes, so pure ethanol cannot be obtained by ordinary distillation. The remaining water is removed by a chemical drying method or by azeotropic distillation with an entrainer.

Hinglish: Azeotrope aisa mixture hai jiska vapour aur liquid same composition rakhhte hain; isliye fixed temperature par boil hota hai aur composition nahi badalti. Ethanol aur paani minimum boiling azeotrope banate hain, lagbhag 95 percent ethanol par. Fractional distillation is composition tak hi kaam karti hai, uske baad distillate same rehta hai, isliye pure ethanol seedhe distillation se nahi milta.

Q7 Calculate the boiling point of (i) a 1 molal aqueous glucose solution and (ii) a 1 molal aqueous calcium chloride solution, assuming complete dissociation. K_b for water is 0.52 K kg mol^-1.

Answer (English): Glucose does not dissociate, so i is 1 and the elevation is 1 times 0.52 times 1, which is 0.52 K. The normal boiling point of water is 373.15 K, so the solution boils at 373.67 K. Calcium chloride gives three ions, Ca2+ and two Cl-, so i is 3 and the elevation is 3 times 0.52 times 1, which is 1.56 K. That solution boils at 374.71 K. The electrolyte produces a much larger elevation because the number of particles is three times as large for the same formula concentration.

Hinglish: Glucose ke liye i = 1, isliye ΔTb = 0.52 K aur boiling point 373.67 K. CaCl2 teen ions deta hai, i = 3, isliye ΔTb = 1.56 K aur boiling point 374.71 K. Same molality par electrolyte ka effect teen guna zyada hota hai kyunki particles zyada bante hain.

Q8 Glucose weighing 1.8 g is stirred into 100 g of water. Find the freezing point of the resulting solution. K_f for water is 1.86 K kg mol^-1 and the normal freezing point of water is 273.15 K.

Answer (English): Glucose is 1.8 divided by 180, or 0.010 mol. The solvent mass is 0.100 kg, so the molality is 0.100 mol kg^-1. Glucose is a non-electrolyte, so i is 1 and the depression is 1.86 times 0.100, which is 0.186 K. The solution freezes at 273.15 minus 0.186, which is 272.96 K, and on the Celsius scale that is about minus 0.19 degrees. The freezing point falls because the vapour pressure of the solution is lower than that of pure water.

Hinglish: 1.8 g glucose = 0.010 mol, aur 100 g paani = 0.100 kg, isliye molality 0.100 m hai. Glucose non-electrolyte hai, i = 1. ΔTf = 1.86 × 0.100 = 0.186 K, isliye freezing point 273.15 - 0.186 = 272.96 K hoga. Solution ka freezing point hamesha pure solvent se kam hota hai.

Q9 Define osmotic pressure. Calculate the osmotic pressure of a 0.1 M glucose solution at 300 K and explain what an isotonic solution is.

Answer (English): Osmotic pressure is the extra pressure needed on the solution side to hold back solvent that would otherwise cross a semipermeable membrane into it. For glucose i is 1, so the pressure is C R T with C equal to 0.1 mol L^-1, R equal to 0.0821 L bar K^-1 mol^-1 and T equal to 300 K. The product is 2.463 bar. Two solutions are isotonic when they have the same osmotic pressure, so no net movement of solvent occurs across a membrane between them. A 0.9 percent sodium chloride solution is isotonic with blood, which is why it is used for intravenous drips; a more concentrated solution would draw water out of red blood cells and a more dilute one would push water in.

Hinglish: Osmotic pressure wo extra pressure hai jo solution par lagana padta hai taaki semipermeable membrane se solvent ka flow ruk jaaye. Glucose ke liye Π = CRT = 0.1 × 0.0821 × 300 = 2.463 bar. Jab do solutions ka osmotic pressure same hota hai to unhe isotonic kehte hain; 0.9 percent NaCl blood ke saath isotonic hota hai isliye IV drip mein use hota hai.

Q10 Potassium sulphate dissociates completely in water. Find the van't Hoff factor and the depression in freezing point for a 0.1 molal solution. K_f for water is 1.86 K kg mol^-1.

Answer (English): One formula unit of K2SO4 gives two potassium ions and one sulphate ion, so n is 3 and complete dissociation gives i equal to 3. The depression is i times K_f times m, which is 3 times 1.86 times 0.1, giving 0.558 K. The freezing point of the solution is therefore 273.15 minus 0.558, or 272.59 K. The measured i in a real solution is slightly less than 3 because ion pairing reduces the number of free particles, so the measured depression is slightly smaller than this calculated value.

Hinglish: K2SO4 teen ions deta hai (2K+ aur SO4^2-), isliye complete dissociation par i = 3. ΔTf = i K_f m = 3 × 1.86 × 0.1 = 0.558 K, aur freezing point 272.59 K. Real solution mein ion pairing ki wajah se i thoda 3 se kam hota hai, isliye actual depression bhi thoda kam aata hai.

The five question quiz for this chapter is attached to this page. It covers concentration units, Henry's law, Raoult's law and the colligative properties, and each answer comes with an explanation. Take the quiz after you have written out the ten answers above, because the quiz checks the same ideas with different numbers.