Title Med-RM_Chem_SP-2_Ch-10-Solutions.pdf ✅

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Chapter Contents Introduction Solution is a homogenous mixture of two or more components. Solvent is the component whose physical state is the same as that of solution. In case both the components are in same phase, solvent is the component which is in larger proportions. Types of solutions A solution can be solid, liquid or a gas depending upon the physical state of the components. The various types of solutions are tabulated below. S. No. Solute Solvent Types of Solution Examples 1. Solid Solid Solid in Solid Alloys 2. Liquid Solid Liquid in Solid Hydrated salts 3. Gas Solid Gas in Solid Dissolved gases in minerals 4. Solid Liquid Solid in Liquid Salt solution in water 5. Liquid Liquid Liquid in Liquid Alcohol in water 6. Gas Liquid Gas in Liquid Aerated drinks 7. Solid Gas Solid in Gas Iodine vapours in air 8. Liquid Gas Liquid in Gas Humidity in air 9. Gas Gas Gas in Gas Air CONCENTRATION TERMS S.No. Concentration Term Formula Definition 1. Mole fraction BA A A nn n +  2. Mass percent Mass % of A = 100 × + BA A ww w 3. Volume percent Volume % of A = 100 × + BA A VV V The ratio of the number of moles of one component to the total number of moles of all the components present in the solution. The number of parts by mass of one component (solute or solvent) per 100 parts by mass of solution. The number of parts by volume of one component (solute or solvent) per 100 parts by volume of the solution. Introduction Concentration Terms Solubility of Gases Vapour Pressure of Solutions of Solids in Liquids Azeotropic Mixture Colligative properties Chapter 10 Solutions
102 Solutions NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 S.No. Concentration Term Definition Formula 4. Parts per million (ppm) ppm = 106 Massof solution Massof × A The number of parts by mass (or by volume) of one component per million parts by mass (or by volume of solution). 5. Molarity (M) M = Volume of solution in litres Gram molesof solute The number of gram moles of the solute dissolved per litre (or dm )3 of the solution. 6. Normality (N) N = Volume of solution litresin Gram equivalent sof solute The number of gram equivalents of the solute dissolved per litre of the solution. 7. Molality (m) m = Massof kgin solution Gram molesof the solute The number of gram moles of solute dissolved in 1000 gm of solvent. Some Important Relations : Molarity on mixing solutions of same solute M1 V1 + M2 V2 = M3 (V1 + V2 ) Normality on mixing solutions of same solute N1 V1 + N2 V2 = N3 (V1 + V2 ) Relation between Molarity and density (g/cc) Molarity (M) = Percentage strength (w/w) Density (gm/ml) 10   Molar mass of solute Relation between normality and density (g/cc) Normality (N) = Percentage strength (w/w) Density (gm/ml   ) 10 Equivalent mass of solute Relation between Normality and Molarity : Normality = n factor × molarity Relation between molarity and molality : Molarity 100 Molality 1000 density Molarity molar mass of solute     PPM = mass fraction × 106 SOLUBILITY OF GASES Gases can diffuse with each other similarly they dissolve in liquids and solids. The solubility of gas in a liquid is determined by several factors like temperature and pressure. The solubility of a gas in a liquid is given by Henry's Law, according to which solubility of gas in a liquid is directly proportional to the pressure of gas. The solubility of gases depends on the partial pressure of gas and mole fraction of the gas in the solution is proportional to the partial pressure of the gas in vapour phase hence Partial pressure of the gas K mole fraction of the gas   H i.e. p = KH × , where KH is Henry’s constant and is the function of nature of the gas.
NEET Solutions 103 Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Note : 1. Higher the value of KH at a given pressure, lower is the solubility of the gas in the liquid. 2. With increase in temperature solubility decreases. 3. Plot of partial pressure of the gas versus mole fraction of gas in solution is a straight line and slope of the straight line is equal to Henry’s Constant (KH). Example 1 : Henry's law constant of CO2 in water at 298 K is 5 3 k bar. If pressure of CO2 is 0.01 bar, find its mole fraction. Solution : p = KH  X  CO H (CO ) 2(g) 2 p KX    CO2 5 0.01 1000 X 3    2 6 X 6 10 CO    Vapour Pressure The pressure exerted by the vapours in equilibrium with its liquid at a given temperature. Factors on which vapour pressure depends : Nature of liquid : When the intermolecular forces of attractions are stronger then the vapour pressure will be low because less number of molecules can leave the liquid. e.g. 25 25 o pC H OC H  2 5 o pC H OH Temperature : Higher the temperature, greater would be the vapour pressure. This is because when temperature is raised, kinetic energy of the molecules increases and therefore more number of molecules leave the surface of the liquid Raoult's Law In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in the pure state. Let us consider a mixture of two completely miscible volatile liquids A and B, having the mole fraction A and B. Suppose at a certain temperature, their partial vapour pressures are pA and pB and the vapour pressures in the pure state are p°A and p°B. According to the Raoult's Law, pA = Ap°A and pB = Bp°B ...(i) If P is the total pressure of the system at the same temperature, then P = pA + pB = Ap°A + Bp°B ...(ii) or P = (1 – B)p°A + Bp°B = (p°B - p°A)B + p°A ...(iii) Since p°A and p°B are constant at a particular temperature, it is evident from eqn. (iii) that the total vapour pressure is a linear function of the mole fraction B(or A = 1 – B). Examples : Benzene + Toluene, Ethanol + Methanol, and CH3 CHO + C2 H5 CHO etc. Mole Fraction of a Component in the Vapour Phase:   0 p PY A A TA and   0 p PY BB T B . Mole fraction in vapour phase, 0 A A A 0 0 A A BB p (Y ) p p     .
104 Solutions NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Example 2 : Vapour pressure of CH3Cl and CH2Cl2 are 540 mm Hg and 402 mm Hg respectively. 101 g of CH3Cl and 85 g of CH2Cl2 are mixed together. Determine (i) The pressure at which the solution starts boiling. (ii) Molar ratio of solute v/s solvent in vapour phase in equilibrium with solution. Solution : (i) Boiling occurs when external pressure becomes equal to the vapour pressure. So, the boiling pressure = V.P. of solution px px A A BB     Let A = CH3Cl, B = CH2Cl2, then Toal pressure 2 1 540 402 3 3    = 360 + 134 = 494 mm Hg (ii) Here the solute is CH2Cl2 as mass is less CH Cl 2 2 1 402 3 134 x 494 494    CH Cl 3 2 540 3 360 x 494 494    Now, 2 2 3 CH Cl solute(g) CHCl solvent(g) n n 134 0.372 n n 360     VAPOUR PRESSURE OF SOLUTIONS OF SOLIDS IN LIQUIDS If a non-volatile solute is added to a solvent to give a solution, the vapour pressure of solution is solely from the solvent alone. This vapour pressure of the solution at a given temperature is found to be lower than the vapour pressure of the pure solvent at the same temperature. According to Raoults's law if p is vapour pressure of the solvent, 1 be its mole fraction, p1  be its vapour pressure in the pure state, Then, 0 p° solvent 1 Vapour pressure p1  1 and p1 = 1 1p  The proportionality constant is equal to the vapour pressure of pure solvent, p1  . A plot between the vapour pressure and the mole fraction of the solvent is linear. Ideal Solutions The solution in which solute-solute interaction in pure state and solvent-solvent interaction in pure state are almost similar to the solute-solvent interactions in solution. H(mix) = 0 po B p = p + p A B pA pB B po A B A = 0  = 1   B A = 1 = 0 Vapour pressure V(mix) = 0 S(mix) > 0 G(mix) < 0 pA = poA × A and pB = poB × B.