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At 100°C and 1 atm, if the density of liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (2000S)
  • a)
    6 cm3
  • b)
    60 cm3
  • c)
    0.6 cm3
  • d)
    0.06 cm3
Correct answer is option 'C'. Can you explain this answer?
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At 100°C and 1 atm, if the density of liquid water is 1.0 g cm&nda...
Mass of 1 L of vapour = volume × density                    
= 1000 × 0.0006 = 0.6 g
V of liquid water = 
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At 100°C and 1 atm, if the density of liquid water is 1.0 g cm&nda...
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Ideal Solution at Fixed TemperatureConsider two liquids B and C that form an ideal solution. We hold the temperature fixed at some value T thatis above the freezing points of B and C. We shall plot the systems pressure P against XB, the overall molefraction of B in the system :Where are the number of moles of B in the liquid and vapor phases, respectively. For a closed system XB is fixed, although may vary.Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath. To see what the P-versus–XB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figur e) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition at D is equal to the overall mole fraction XB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoults law, relates the vapour-phase mole fractions to the liquid composition as follows :............(1)Where PB0 and PC0 are the vapour pressures of pure B and pure C at T, where the systems pressure P equals the sum PB + PC of the partial pressures, where, and the vapour is assumed ideal. ............(2)Let B be the more volatile component, meaning that . Above equation then shows that The vapour above an ideal solution is richer than the liquid in the more volatile component. Equations (1) and (2) apply at any pressure where liquid –vapour equilibrium exists, not just at point D.Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, wereach point F in figure , where the last drop of liquid vaporizes. Below F, we have only vapour. For any pointon the line between D and F liquid and vapour phases coexist in equilibrium.Q. Two liquids A and B have the same molecular weight and form an ideal solution. The solution has a vapour pressure of 700 Torrs at 80ºC. It is distilled till 2/3rd of the solution (2/3rd moles out of total moles) is collected as condensate. The composition of the condensate is xA = 0.75 and that of the residue is xA= 0.30. If the vapour pressure of the residue at 80ºC is 600 Torrs, find the original composition of the liquid ?

Ideal Solution at Fixed TemperatureConsider two liquids B and C that form an ideal solution. We hold the temperature fixed at some value T thatis above the freezing points of B and C. We shall plot the systems pressure P against XB, the overall molefraction of B in the system :Where are the number of moles of B in the liquid and vapor phases, respectively. For a closed system XB is fixed, although may vary.Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath. To see what the P-versus–XB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figur e) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition at D is equal to the overall mole fraction XB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoults law, relates the vapour-phase mole fractions to the liquid composition as follows :............(1)Where PB0 and PC0 are the vapour pressures of pure B and pure C at T, where the systems pressure P equals the sum PB + PC of the partial pressures, where, and the vapour is assumed ideal. ............(2)Let B be the more volatile component, meaning that . Above equation then shows that The vapour above an ideal solution is richer than the liquid in the more volatile component. Equations (1) and (2) apply at any pressure where liquid –vapour equilibrium exists, not just at point D.Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, wereach point F in figure , where the last drop of liquid vaporizes. Below F, we have only vapour. For any pointon the line between D and F liquid and vapour phases coexist in equilibrium.Q.If the above process is repeated for all other compositions of mixture of C and B. If all the points where vapours start converting into liquid are connected and all the points where vapours get completely converted into liquid are connected then obtained graph will look like

In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .

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At 100°C and 1 atm, if the density of liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (2000S)a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer?
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At 100°C and 1 atm, if the density of liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (2000S)a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 100°C and 1 atm, if the density of liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (2000S)a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 100°C and 1 atm, if the density of liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (2000S)a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer?.
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