Solutions: Solubility of a Gas in a Liquid (Henry's Law) ‘It states that, at a constant temperature, the solubility of a gas in liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution. pes or p=Ky.S where, 5 = solubility of gas partial pressure of gas at equilibrium Ky =Henry’s law constant. Unit of solubility is same as that of concentration, 1 Mark Questions Multiple Choice Questions 1, Value of Henry's constant Ky CBSE 2023 (a) increases with decrease in temperature (0) decreases with increase in temperature (©) increases with increase in temperature (@) remains constant 2. Solubility of gases in liquids decreases with rise in temperature because dissolution is an (CBSE SQP 2021-22 (Term) (@) endothermic and reversible process (©) exothermic and reversible process (©) endothermic and irreversible process {(@) exothermic and irreversible process Pressure does not have any significant effect on solubility of solids in liquids because (@)solids are highly compressible CBSE 2021-22 (Term-t) (liquids are highly compressible (©)solubility of solid in liquid is directly proportional to partial pressure (@)solids and liquids are highly incompressible Assertion-Reason Directions (Q. Nos. 3 and 4) In the following questions an Assertion (A) is followed by a corresponding Reason( R). ‘Use the following keys to choose the appropriate answer. (@) Both (A) and (R) are correct and (R) is the correct explanation of (A). (b) Both (A) and (R) are correct but (R) is not the correct explanation of (A). (©) (A)is correct and (R) is incorrect. (@) (A) is incorrect and (R) is correct. STS RRTERE OT ars’ Questions” « Inother words, “the partial pressure of the gas in vapour phase (p) is directly proportional to the mole fraction (y.) of the gas in the solution. pey or p=Ky-% «+ Ifwe draw a graph between partial pressure of the gas versus mole fraction of the gas in solution, we get straight line whose slope is given by Ky. © Higher the value of Ky at a given pressure, lower is the solubility of the gas in the liquid, Solubility of gases increases with increase in pressure. 4, Assertion (A) Molarity of a solution changes with temperature. Reason (R) Molarity is a colligative property. CBSE SQP 2023, 5. Assertion (A) Aquatic species are more comfortable in cold water rather than in warm water. Reason (R) Different gases have different Ky, values at the same temperature. ‘CBSE SP 2021 Very Short Questions 6. Give reason ‘Aquatic animals are more comfortable in cold water than in warm water. (CBSE 2024, 2018 C 7. Define the term mole fraction. ‘All India 2012, 2010C, Delhi 2012 8. Explain the Henry's law about dissolution of a gas in aliquid. All india 2012, 2011; Delhi 2011; Foreign 2011 9. State the main advantage of molality over molarity as the unit of concentration. Dethi 2010 2 Marks Questions 10. State Henry's law. Calculate the solubility of CO, in water at 298 K under 760 mm He. (Ky for CO, in water at 298 K is 1.25% 10° mm Hg) All Indio 2020 11. Calculate the molality of ethanol solution in which the mole fraction of water is 0.88. CBSE SqP 2019 12, Give reason for the following. (i) Aquatic species are more comfortable in cold ‘water than in warm water. ii) At higher altitudes people suffer from anoxia resulting in inability to think, All india 201913. Define the following terms. (i) Mole fraction (1) (i) Molality ofa solution (m) 14, Calculate the molarity of 9.8% ( 1,80, if the density of the solution [Molar mass of HO, =98 8 mol] 45. Differentiate between molarity and molality of a solution, How can we change molality value of a solution into molarity value? Delhi 2014C; Foreign 2011 16. A solution of glucose (C,H,30g) in water is labelled ‘as 10% by weight. What would be the molality of the solution? (Molar mass of glucose = 180 g mol” ). All Al India 2015C {v) solution of 1.02. g mL“! Foreign 2014 2013 17. Ifthe density of water of a lake is 1.25 gmL"' and 1 kg of lake water contains 92 g of Na° ions, calculate the molarity of Na” ions in this lake water. (Atomic mass of Na=23 gmol"'). Foreign 2012 ROA KEY [7 « Find volume of solution using the formula V- || © While calculating volume of solution, total mass(m), ie. mass of solute + mass of solvent is taken. 18, Differentiate between molarity and molality for a solution. How does a change in temperature influence their values? Dethi 2011, Foreign 2011 19, Explain why aquatic species are more comfortable in cold water rather than in warm water. Delhi 20126 > ~>s swuestion Bank : CHEMIST, 20. State Henry's law and mention its two im Portant applications. Al Inia 2 a 2024, 25 ¢ 21, Answer the following questions, cose SP 2m) (0 State Henry's law and explain why the tank by aba divers are filled wit ar diluted wi iclium (11.7% helium, 56.2% ae nitrogen and 32.1% (ii) Assume that argon exertsa partial pressure of 6 bar. Calculate the solubility of argon gas in water. (Given Henry's constant for argon dssoley in water, Ky, =40 kbar) 3 Marks Questions 22. A solution of glucose (molar mass = 180 g mel") in water is labelled as 10% (by mass). What would be the molality and molarity ofthe solution? (Density of solution = 1.2 g mL~). Delhi 2014 23, The partial pressure of ethane over a saturated solution containing 656 x 10~ g of ethane is 1 bar. If the solution contains 50x 10~g of ethane, then what will be the partial pressure of the gas? Delhi 2013C; Al inc 2012¢ 24, IFN, gas is bubbled through water at 293 K, how many millimoles of N, gas would dissolve in 1 L of water? Assume that N exerts a partial pressure of 0.987 bar. Given that Henry's Jaw constant for N, at 293 K is 76.48 K bar. All India 20126 1. (©) increases with increase in temperature. w 2. (b) The dissolution ofa gas in a liquid is an ‘exothermic process. So, heat is evolved. ‘On applying Le-Chatelier’s principle the increase of ‘temperature decreases the solubility, Gas + Solvent Solu (Cacia it swans ‘eka dieting). n+ Heat 3 (2) According {0 Henry's law, solubility of gas ina id is directly proportional to pressu gas above the surface of solution, ne OF Mt The solubility of wi ‘eases as expected, increases wi increase in pressure, " eee ta 4, (c) Assertion is correct, but Reason is incorrect. Molarity is not a colligative property. It is only & concentration term. It does not depend on number of solute particles. 5. (b) Both (A) and (R) are correct but (R) is not the correct explanation of (A). Aquatic species are more comfortable in cold water rather than in warm water because the solubility of gas with increase of temperature. issolved state in water. As per Henry's lay, when temperature rises solubility of & gas decreases in solvent, it means solubility of oxygen jess than cold water. This makes respirate comfortably in cold water. decrea:Solutions 7. The mole fraction of a component is the ratio of the number of moles of the component tothe total number of moles of all the components present in the solution Mole fraction of a component of the component Total number of moles of all the components Fora binary solution, mole fraction of component 4, ™ Xa m4 + Similarly, for B, x5 =—"®—and x4 +% 5 n+p 8, Henry’s law states that, the partial pressure of the gas in vapour phase ( p) is directly proportional to the mole fraction of the gas ()in the solution. paKy x Here, Ky = Henry's law constant, Different gases have different Ky, values at the same temperature. 8. Molality does not change with change in temperature while, molarity decreases with rise in temperature. 10. According to Henry's law, p=Ky °% 08% 107% umber of moles of ethanol, umber of moles of water. Molality of ethanol means the number of moles of ‘ethanol present in 1000 g of water. rn, =1008 55,5 moles 18 : ‘Substituting the value of nm, in Eq. (i), we get ma 0.12 55.5 +m, ny = 7.57 moles ‘Thus, molality of ethanol (Cp11 sOH) =7.57 m 12, (j) Refer to sol 5. o i) At higher altitudes, people suffer from anoxia because at higher altitudes, the partial pressure of oxygen is less than that at ground level. This leads to low concentration of oxygen in the blood and tissues of people. o 13, (i) Mole fraction (7) Refer to texton page 2. (1y (ii) Molality of solution (m1) Refer to text on page 2.¢1) 0.8% solution of H SO, means 98 g of 11,50 q is present in 100 g of the solution, Thus, Mass of 11,50, dissolved = 98 Mass of solution = 1002 Density of solution = 102 g mL Moles of H,S0, = 9804 mL. or 0.09804 L. of moles of the solute ‘Volume of solution (in litres) 0.1 mol 0.09804 L =1019 mol L“ or 1.02 M 15, Molality is defined as the number of moles of the solute per kilogram of the solvent. It is represented by m. Number of moles of solute «1000 ‘Mass of solvent (ing) Motality (m) = It does not change with change in temperature. Molarity is defined as the number of moles of solute dissolved in one litre or one cubie decimetre of the solution, _Nunber of moles of slut 1000 ; nm ex Molarty (8) = tums of solution (in iL) Ttdecreases with increase in temperature (as V 7). We ean change molality value ofa solution into molarity value by using following relation. Mx 1000, (1000% d= ( Where, Af is the molarity and Mf is the molar mass of Component 2 (generally solute) and d isthe density of solution (in gem *). 16. A 10% glucose solution by weight means that 10 g of iglucose is present in 100 g solution. Molality (1) = Number of moles of 10 g glucose “a 0555 mol ‘Weight of glucose = 10g, 90 Weight of water = 90g = oa kg =009kgyy Number of motes of solute Mass of solvent (in kg) = 20555 100 «961 mol kg! or 061m O09kg Molaity (m) = 17. Given, d= 1.25 gm", w, =92g, W, =1 kg or 1000 g, Mf, =23 g mol! We know that, density (d) of solution Mass (mi) of solution Volume (7) of solution im _ (1000 g+ 92 8) a 125em4 mass of solvent + mass of solute] or [Mass of solution 312 int 28736 mL =8736%107 L 1.25 28 23.gmol Moles of solute Moles of solute = mol 4 mol > Chapterwise CBSE Question Bank : CHEMIsy, ey where, Kyy= Henry's constant %=mole fraction ofthe gas Sx1ot 40x10 22. 10% glucose solution (by mass) in water m of glucose (solute is present in 100g of solution Thus, mass of solution, m=100 g wz =10g (solute); H, =90g (solvent) My (glucose) = 180 g mol! Density of solution = 1.2 g mL“! Volume of solution, yom. 100g 1000 12g mL- ; W 10x12 Molarity (4) =——"2_ = 01? _ cams M,x volume 180x1 aa Molality (m) = 12% 1000 _ 10% 1000 _ 9 61755 M,xW,— 180x90 Molarity =. = Volume of solution (L) 8736x107 L. =4579mol L™ or 4.58M 18. Refer to text on page 2 and solution 15. 19. Refer to solution 5. 20. Henry’s law Refer to text on page 3, Applications of Henry’s Law (@ To increase the solubility of CO, in softdrinks and soda water, the bottle is sealed under high pressure. Gi) To minimise the painful effects of bends or decompression sickness in deep sea divers, oxygen diluted with less soluble helium gas is used as breathing gas. (1/24 1/2=1) 21. (i) Henry's law states that the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution. The pressure under the water is very high. So, the solubility of gases in blood increases. When the diver comes to surface, the pressure decreases. ‘The releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood, this ccan creates medical issues. To avoid this situation and maintain the same partial pressure of nitrogen underwater too, dilution is done with helium. Gi) Henry's law is given by formula paKyx w Common eneraly, students write wrong uri fr Mistake molarity and molaity. Do wet the correct units atthe end of the answer you get. 23. According to Henry’s law, Hx P =Ky xIbar 56x 10"2g bar! m InI case, 6.56% 10> or Ky In Il case, $.00x 107? g = (656x107 g bar“')x p 5.00x107g 6.56% 107 g bar 24, According to Henry’s law, py, =Ku *%ny or 0.762 bar eye, = PND = 0987 bat 4.29107 N2 "Ky 76480 bar If mmoles of N, are present in 1L or 1000 g of water (« 10005 n N+55,5 55.5 1.29 107% (sn <<< 55.5) 129% 10° x55,5 moles 71,395%10™ moles =0.716 millimoles