SOLUTIONS Topic-2 RAOULT S LAW, ALICATIONS AND NUMERICALS VERY SHORT ANSWER QUESTIONS 1. Define vapour preure? An: When a liquid i in equilibrium with it own vapour the preure exerted by the vapour on the urface of the liquid i known a the vapour preure of the liquid. 2. Stat Raoult Law? An: For a olution containing non volatile olute, the relative lowering of vapour equal to mole fraction of olute. X B preure i na + Simplified (or) reduced form of Raoult law: 3. What prediction can you make for relation between vapour preure and temperature? An: The vapour preure of the liquid i directly proportional to the temperature of the liquid. The vapour preure of a liquid i independent of hape of the veel. Vapour preure of liquid increae exponentially with increae in temperature. 4. Solution A, B, C, D have vapour preure 3mm, 4mm, 6mm, and 8mm of Hg at 25ºC repectively. Which i the mot volatile and why? An: D i the mot volatile becaue of high vapour preure. Thi i due the effect of non-volatile on the vapour preure of the the liquid. 5. What i boiling point of liquid? How it i effected by external preure? An: The temperature at which the vapour preure of the liquid i equal to the atmopheric preure i known a the boiling point of the liquid. Boiling point of a liquid can be changed by changing the external preure. If external preure i increaed, the boiling point of a liquid i increaed and vice-vera. 6. Write any two limitation of Raoult Law? An: Raoult Law i applicable to; 1. Dilute olution only
2. The olute mut be non-volatile. 3. Solute which doe not undergo in it molecular weight. 4. to ideal olution. 7. What i an ideal olution? An: The olution which obey Raoult law at all concentration of temperature are called ideal olution. In cae of ideal olution, Vmixing and Hmixing. 8. What happen to the magnitude of the boiling point of a olvent if a non-volatile olute i diolved in it? An: The vapour preure of olution i le than that of the olvent. Therefore the Temperature required to raie the vapour preure of the olution to the atmopheric preure i greater than the temperature required to raie the vapour preure of the olvent to the value equal to the atmopheric preure. Hence, the boiling point of olution will be greater than that of the olvent. SHORT ANSWER QUESTIONS 1. What i vapour preure? How doe it vary with temperature? An: When a liquid i in equilibrium with it own vapour the preure exerted by the vapour on the urface of the liquid i known a the vapour preure of the liquid. The vapour preure of the liquid mut be called a aturated vapour preure, becaue actually the atmophere over the liquid, which i aturated with the vapour of the liquid, exert the preure on the liquid. The vapour preure of the liquid i repreented by. The vapour preure of the liquid i directly proportional to the temperature of the liquid. Vapour preure of liquid increae exponentially with increae in temperature. Log V T 1 give a traight line with ve lope. 2. State and explain Raoult law? Raoult law: For a olution containing non volatile olute, the relative lowering of vapour preure i equal to mole fraction of olute. X B na + Simplified (or) reduced form of Raoult law: n n B A (for dilute olution, i very mall and it can be neglected) w M m W Where, Vapour preure of pure olvent Vapour preure of olution
X B mole fraction of olute m molecular weight of olute M molecular weight of olvent w weight of olute W weight of olvent 3. What i the relation between the vapour preure and it boiling point? An: The vapour preure of a liquid increae with increae in temperature. Thi goe on until the critical temperature of the liquid i reached. Above the critical temperature liquid tate doe not exit when the vapour preure of the liquid become equal to the external (atmopheric) preure the liquid i aid to be boiling and the temperature at which thi happen i known a boiling point. Water boiling point i 1ºC at 1 atm. reure. If the external preure i reduced, the liquid boil at lower temperature. The boiling point of liquid can be increaed by increaing the external preure. 4. What i the role of non-volatile olute in lowering of vapour preure of olvent? An: When a non-volatile olute i added to a olvent, the vapour preure i lowered due to the following reaon: ercentage urface area occupied by the olvent decreae. Thu the rate of evaporation and vapour preure decreae. The olute molecule occupy the urface, and o the per cent urface area occupied by the olvent decreae. According to Graham law of evaporation, Rate of evaporation 1/ denity When a non-volatile olute i diolved in a liquid, it denity increae. Thu both rate of evaporation and vapour preure are lowered. 5. What i relative lowering of vapour preure? How i it ueful in determining the molecular weight of a olute? An: When a non-volatile olute i diolved in a liquid, it denity increae. Thu both rate of evaporation and vapour preure are lowered. If p i the vapour preure of pure olvent and p i the vapour preure of the olution, The difference (p p ) i termed lowering in vapour preure, and the ratio [p p /p ] i termed relative lowering in vapour preure.
Raoult, etablihed a relationhip between relative lowering in vapour preure and compoition of the olution after a erie of experiment in variou olvent. Uing thi relationhip one can determine the molecular weight of olute by uing thi following equation. w M m W Where, Vapour preure of pure olvent Vapour preure of olution X B mole fraction of olute m molecular weight of olute M molecular weight of olvent w weight of olute W weight of olvent LONG ANSWER QUESTIONS 1. What do you mean by vapour preure of a liquid? Decribe the vapoiuriation and condenation proce in a cloed veel. How doe vapour preure change with preure? An: The preure exerted by the vapour of a liquid when it i in equilbrium with liquid i called vapour preure. The proce of tranformation of liquid into vapour i called evaporation and the proce of tranformation of vapour into liquic i called condenation. In a lquid preent in cloed veel rate of evaporation i equal to rate of condenation. Thi tage i called equilibrium tage and the equilibrium i called dynamic equilibrium. Effect of temperature: A you know, liquid will evaporate. The rate and extent to which it evaporate depend on the temperature. With rie in temperature of a liquid the average kinetic energy of liquid molecule increae. Thi increae in kinetic energy overcome the attractive force between molecule and hence molecule ecape more rapidly into air. Thu the vapour preure of a liquid increae with increae in temperature. The increae in vpour preure of a liquid with rie in temperature in non linear but it i exponential.
The amount of evaporation increae when the temperature increae. When the temperature i uch that the vapor preure i jut a high a the atmopheric preure, the liquid boil. That temperature i called the boiling point. 2. Define and explain Raoult law. How i it ueful in determining molecular weight of a non-volatile olute? An: Raoult law: For a olution containing non volatile olute, the relative lowering of vapour preure i equal to mole fraction of olute. X B -------------------(1) na + Simplified (or) reduced form of Raoult law i given by; n n B A (for dilute olution, i very mall and it can be neglected) w M -----------------------(2) m W
From equation (2) molecular weight of olute (m) can obtained by the following equation. Where, p o. w. M m --------------------------- (p o p ) W Vapour preure of pure olvent Vapour preure of olution X B mole fraction of olute m molecular weight of olute M molecular weight of olvent w weight of olute W weight of olvent Raoult law i applicable to : Ideal olution dilute olution olution containing non volatile olute no change in the interaction before and after mixing of liquid component in cae of olution containing micible liquid. Solute which neither diociate nor aociate. 3. Define Raoult law for lowering of vapour preure. How i the law ueful in determing the molecular weight of olvent when a known non-volatile olute i diolved? An: Raoult law: For a olution containing non volatile olute, the relative lowering of vapour preure i equal to mole fraction of olute. X B -------------------(1) na + Where, n B and n A are number of mole of olute and olvent repectively.
Simplified (or) reduced form of equation (1) i given by; --------------------------(2) na (for dilute olution, n B i very mall and it can be neglected) Where, Vapour preure of pure olvent Vapour preure of olution X B mole fraction of olute We know that n B w/m and n A W/M where, m molecular weight of olute M molecular weight of olvent w weight of olute W weight of olvent By ubtituting thee value in equqtion(2). w M ---------------- (3) m W Form the above equation the molecular weight of olvent (M) i given by; m. (p o p ). W M -------------------------- p o. w Example: 1 NUMERICALS Calculate the vapour preure lowering caued by addition of 5 g of ucroe (molecular ma 342) to 5 g of water if the vapour preure of pure water at 25 C i 23.8 mm Hg. According to Raoult law, p p /p n/n+n
or p ( n/n+n)p Given n 5/342.416, N 5/18 27.78 and p 23.8 Subtituting the value in the above equation, Example: 2 p (.146/.146+27.78) 23.8.124 mm Hg The vapour preure of pure benzene at a certain temperature i 64mm Hg. A nonvolatile olid weighing 2.175 g i added to 39. g of benzene. The vapour preure of the olution i 6 mm Hg. What i the molecular ma of the olid ubtance? Solution: According to Raoult law. (p p )/p n/(n+n ) Let m be the molecular ma of the olid ubtance. n 2.175/m ; N 39/78.5 [Molecular ma of benzene 78] Subtituting the value in the equation. p o. w. M m --------------------------- (p o p ) W 64. 2.175. 78 m --------------------------- (64 6 ) 39 m 65.25 Thu molecular weight of olute i 65.25 Example: 3 A olution containing 3 g of a non-volatile olute in exactly 9 g of water ha a vapour preure of 21.85 mm of Hg at 25 C. Further 18 g of water i then added to the olution; the new vapour preure become 22.15 mm Hg of at 25 C. Calculate (a) molecular ma o the olute and (b) vapour preure of water at 25 C. Solution:
Let the vapour preure of water at 25 C be p and molecular ma of the olute be m. Uing Raoult law in the following form. For olution (I), (p 21.85)/21.85 3 18/9 m (i) For olution (II), (p 22.15)/22.15 3 18/18 m (ii) Dividing Eq. (i) by Eq. (ii), Example:4 (p 21.85)/21.85 22.15/(p 22.15) 18/9 6/5 Subtituting the value of p in Eq. (i) the value of m i given by; M 67.9 What ma of non-volatile olute (urea) need to be diolved in 1 g of water in order to decreae the vapour preure of water by 5%. What will be the molality of olution? Solution: Uing Raoult law in the following form, p p /p wm/wm If p 1 mm, then p 75 mm 1 75/75 w 18/1 6 w 111.1 Molality w 1/m W 111.1 1/6 1 18.52 m