Retro JEE Advanced (Compilation of Last 10Years Questions)IIT JEE (Main Advanced) past year questions with solutions answer keyRetro IIT JEE (Main and Advanced) have questions from previous/past (10 to 13 years) papers of IIT JEE (Main and Advanced)with solutions,The following set of questions contain previous/past (10 to 13 years) questions of IIT JEE (Main and Advanced) of Chapter State of Matter with solutionsPassage For Q. Nos. 1-2 X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Grahams law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer?

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Passage For Q. Nos. 1-2

X and Y are two volatile liquids with molar weights of 10 g mol^-1 and 40 g mol^-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.
The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

Q. The value of d in cm (shown in the figure), as estimated from Graham’s law, is

(2014 Adv, Comprehension Type)

a)
8
b)
12
c)
16
d)
20

Correct answer is option 'C'. Can you explain this answer?

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Passage IIX and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of the tube of length l = 24 cm , as shown in the figure. The tube is filled with an inert gas at 1 atm pressure and temperature of 300 K. Vapours of X and Y react to form a product which is first formed at a distance, d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and two vapours.Q.The value of d in cm (shown in the figur e), as estimated from Grahams law is

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The variables Pressure, Volume, Concentration and Temperature may change the State of Equilibrium. The change is governed by the Le-Chatelier’s principle. The decomposition of NH3(g) can be made spontaneous by increasing the temperature and lowering pressure. In the reaction, removal of any product from the reaction mixture makes the reversible reaction irreversible and therefore, reaction proceeds to completion. Answer the given questions (i) to (iv).Q. Change in free energy for the equilibrium, gaseous reaction, PCl5 → PCl3 + Cl2 on addition of an inert gas at constant pressure and at constant volume is respectively

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Importance of high specific heat capacity of water for life Specific heat capacity of a substance is the amount of heat required to raise the temperature of that substance by 1 K. It is expressed in the units J/ (kg K). A high specific heat of a substance means that a large amount of heat is required to raise the temperature of the substance. Water has the highest known specific heat capacity. Its specific heat capacity is 4.186 K J/ (kg K) i.e. to raise the temperature of 1 kg of water by 1 Kelvin it requires 4.186 KJ of heat. For comparison sake, Copper requires only 385 Joules of heat to raise 1 kilogram of copper by 1 Kelvin. It also interesting to know that the specific heat capacities in two other phases of water (i.e. ice and water vapour) are less than this. High specific heat of water is mainly due to the presence of a large number of hydrogen bonds between molecules of water. On a beach on a sunny day, it is noticed that the sand is often quite hot to walk on, but the water is always cool, even in the shallows. This is because sand has a lower specific heat capacity than that of water. Sand takes less energy to raise the temperature by one degree. Because water has a high specific heat capacity, it requires more energy to raise the temperature by one degree. Since the sun puts same rate of energy on water and sand, which heats up sand more quickly and water more slowly. Water covers around 70% of the Earth's surface and its high specific heat plays a very important role to sustain life in the earth. It is able to absorb a lot of heat without a significant rise in the temperature. When temperatures decrease, the heat which is stored is released, restraining a rapid drop in temperature. The combined effect is the buffering of heat. A relatively constant temperature without sudden rise and drop is essential to sustain life. Hence water is important for life.Q. To raise the temperature of 1 kg of water and 1 kg of copper by 1 Kelvin, 4.186 KJ and 385 Joule of heat are required respectively. Which one will get heated up faster if exposed to sun?

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The variables Pressure, Volume, Concentration and Temperature may change the State of Equilibrium. The change is governed by the Le-Chatelier’s principle. The decomposition of NH3(g) can be made spontaneous by increasing the temperature and lowering pressure. In the reaction, removal of any product from the reaction mixture makes the reversible reaction irreversible and therefore, reaction proceeds to completion. Answer the given questions (i) to (iv).Q. The equilibrium Solid → Liquid → Gas will shift in forward direction when

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Retro JEE Advanced (Compilation of Last 10Years Questions)IIT JEE (Main & Advanced) past year questions with solutions & answer keyRetro IIT JEE (Main and Advanced) have questions from previous/past (10 to 13 years) papers of IIT JEE (Main and Advanced)with solutions,The following set of questions contain previous/past (10 to 13 years) questions of IIT JEE (Main and Advanced) of Chapter "State of Matter" with solutionsPassage For Q. Nos. 1-2 X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer?

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Retro JEE Advanced (Compilation of Last 10Years Questions)IIT JEE (Main & Advanced) past year questions with solutions & answer keyRetro IIT JEE (Main and Advanced) have questions from previous/past (10 to 13 years) papers of IIT JEE (Main and Advanced)with solutions,The following set of questions contain previous/past (10 to 13 years) questions of IIT JEE (Main and Advanced) of Chapter "State of Matter" with solutionsPassage For Q. Nos. 1-2 X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Retro JEE Advanced (Compilation of Last 10Years Questions)IIT JEE (Main & Advanced) past year questions with solutions & answer keyRetro IIT JEE (Main and Advanced) have questions from previous/past (10 to 13 years) papers of IIT JEE (Main and Advanced)with solutions,The following set of questions contain previous/past (10 to 13 years) questions of IIT JEE (Main and Advanced) of Chapter "State of Matter" with solutionsPassage For Q. Nos. 1-2 X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Retro JEE Advanced (Compilation of Last 10Years Questions)IIT JEE (Main & Advanced) past year questions with solutions & answer keyRetro IIT JEE (Main and Advanced) have questions from previous/past (10 to 13 years) papers of IIT JEE (Main and Advanced)with solutions,The following set of questions contain previous/past (10 to 13 years) questions of IIT JEE (Main and Advanced) of Chapter "State of Matter" with solutionsPassage For Q. Nos. 1-2 X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer?.

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Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Retro JEE Advanced (Compilation of Last 10Years Questions)IIT JEE (Main & Advanced) past year questions with solutions & answer keyRetro IIT JEE (Main and Advanced) have questions from previous/past (10 to 13 years) papers of IIT JEE (Main and Advanced)with solutions,The following set of questions contain previous/past (10 to 13 years) questions of IIT JEE (Main and Advanced) of Chapter "State of Matter" with solutionsPassage For Q. Nos. 1-2 X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. 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Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Retro JEE Advanced (Compilation of Last 10Years Questions)IIT JEE (Main & Advanced) past year questions with solutions & answer keyRetro IIT JEE (Main and Advanced) have questions from previous/past (10 to 13 years) papers of IIT JEE (Main and Advanced)with solutions,The following set of questions contain previous/past (10 to 13 years) questions of IIT JEE (Main and Advanced) of Chapter "State of Matter" with solutionsPassage For Q. Nos. 1-2 X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Retro JEE Advanced (Compilation of Last 10Years Questions)IIT JEE (Main & Advanced) past year questions with solutions & answer keyRetro IIT JEE (Main and Advanced) have questions from previous/past (10 to 13 years) papers of IIT JEE (Main and Advanced)with solutions,The following set of questions contain previous/past (10 to 13 years) questions of IIT JEE (Main and Advanced) of Chapter "State of Matter" with solutionsPassage For Q. Nos. 1-2 X and Y are two volatile liquids with molar weights of 10 g mol-1 and 40 g mol-1 respectively. Two cotton plugs, one soaked in X and the other soaked in V, are simultaneously placed at the ends of a tube of length L = 24cm, as shown in the figure.The tube is filled with an inert gas at 1 atm pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and V to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.Q.The value of d in cm (shown in the figure), as estimated from Graham’s law, is(2014 Adv, Comprehension Type)a)8b)12c)16d)20Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.

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