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Question 5 K Cr2O7 can supply oxygen to oxidise pollutants in a water sample If one kilogram of such water sample requires 20 ml of 0 1 MK Cr2O7 H what is the oxygen requirement of that water in ppm units Options a 96 ppm b 48 ppm c 24 ppm d 32 ppm
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Question 5 K Cr2O7 can supply oxygen to oxidise pollutants in a water sample If one kilogram of such water sample requires 20 ml of 0 1 MK Cr2O7 H what is the oxygen requirement of that water in ppm units Options a 96 ppm b 48 ppm c 24 ppm d 32 ppm
5 1 M solution of monobasic acid HX has molar conductivity x Scm mol while molar conductivity of H and X at infinite dilution is 300 Scm mol and 100 Scm mol respectively The 1 M HX solution contains 0 15 M H If this 1 M solution is filled in a conductivity cell of cell constant 0 1 cm the conductance is found to be Y ohm 1 The value of 10Y is
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5 1 M solution of monobasic acid HX has molar conductivity x Scm mol while molar conductivity of H and X at infinite dilution is 300 Scm mol and 100 Scm mol respectively The 1 M HX solution contains 0 15 M H If this 1 M solution is filled in a conductivity cell of cell constant 0 1 cm the conductance is found to be Y ohm 1 The value of 10Y is
Study the following figure and choose the correct options Assuming complete dissociation of electrolyte 0 05 M aq Al2 SO4 3 solution SPM 0 10 Maq MgCl solution a There will be no net moment of any substance across the membrane b MgCl will flow towards the Al2 SO4 3 solution c Al SO4 3 will flow towards the MgCl solution d The osmotic pressure of 0 1 M MgCl is higher than the of 0 05 M Al SO4 3
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Study the following figure and choose the correct options Assuming complete dissociation of electrolyte 0 05 M aq Al2 SO4 3 solution SPM 0 10 Maq MgCl solution a There will be no net moment of any substance across the membrane b MgCl will flow towards the Al2 SO4 3 solution c Al SO4 3 will flow towards the MgCl solution d The osmotic pressure of 0 1 M MgCl is higher than the of 0 05 M Al SO4 3
The self ionization constant for pure formic acid K HCOOH HCOO has been estimated as 10 4 at room temperature What percentage of formic acid molecules in pure formic acid is converted to formate ion Given d H 1 22 g cc 1 0 0185 2 0 0073 3 0 074 4 0 037
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The self ionization constant for pure formic acid K HCOOH HCOO has been estimated as 10 4 at room temperature What percentage of formic acid molecules in pure formic acid is converted to formate ion Given d H 1 22 g cc 1 0 0185 2 0 0073 3 0 074 4 0 037
Which of the following property is widely used to determin the molar masses of proteins Relative lowering in vapour pressure Elevation in boiling point Depression in freezing point Osmotic pressure
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Which of the following property is widely used to determin the molar masses of proteins Relative lowering in vapour pressure Elevation in boiling point Depression in freezing point Osmotic pressure
Equal mass of a solute are dissolved in equal mass of two solvents A and B and formed very dilute solution The relative lowering of vapour pressure for the solution B has twice the relative lowering of vapour pressure for solution A If MA and MB are the molecular masses of MA MB MB 2MA MA 4MB
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Equal mass of a solute are dissolved in equal mass of two solvents A and B and formed very dilute solution The relative lowering of vapour pressure for the solution B has twice the relative lowering of vapour pressure for solution A If MA and MB are the molecular masses of MA MB MB 2MA MA 4MB
2 54 A solution is prepared by mixing 250 ml toluene C Hg and 8 4 g thiophene C H4S Then molality of thiophene in the solution is Given Density of toluene 0 8 g ml Density of thiophene 1 2 g ml Multiply your answer by 10
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2 54 A solution is prepared by mixing 250 ml toluene C Hg and 8 4 g thiophene C H4S Then molality of thiophene in the solution is Given Density of toluene 0 8 g ml Density of thiophene 1 2 g ml Multiply your answer by 10
33 Given the following nine aqueous solutions 1M CH COOH 0 5M NaOH 3 w v Glucose 1 2 3 0 2M NaCl 6 Centimolar sucrose 7 E 0 2M Urea 4 Decimolar H C O4 8 4 w v NaOH 5 0 9 w v Na CO3 9 Select correct option s A Solution 7 has minimum g L of solute B Only two solutions have same g L of solute C On dividing by ten in four solutions g L of solute value is whole number
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33 Given the following nine aqueous solutions 1M CH COOH 0 5M NaOH 3 w v Glucose 1 2 3 0 2M NaCl 6 Centimolar sucrose 7 E 0 2M Urea 4 Decimolar H C O4 8 4 w v NaOH 5 0 9 w v Na CO3 9 Select correct option s A Solution 7 has minimum g L of solute B Only two solutions have same g L of solute C On dividing by ten in four solutions g L of solute value is whole number
Assertion A Hmix and AVmix are zero for the ideal solution Reason The interactions between the particles of the components of a solution are almost identical between particles in the pure liquids
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Assertion A Hmix and AVmix are zero for the ideal solution Reason The interactions between the particles of the components of a solution are almost identical between particles in the pure liquids
PT P PA B PB 1 Z X Y 0 X Mole fraction of liquid A in liquid solution Y Mole fraction of vapour A in vapour mixture Z is the graphical mid point between 1 X Find value of P Y Given PT 450 mm Hg 100X If P A 300 mm of Hg no 700
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PT P PA B PB 1 Z X Y 0 X Mole fraction of liquid A in liquid solution Y Mole fraction of vapour A in vapour mixture Z is the graphical mid point between 1 X Find value of P Y Given PT 450 mm Hg 100X If P A 300 mm of Hg no 700
Problem1 If N gas is bubbled through water at 293 K how many millimoles of N dissolve in 1 litre of water Assume that N exerts a partial pressure of 0 987 bar Given that Henry s law constant for N at 293 K is 76 48 kbar NCERT Solved Example 097 har
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Problem1 If N gas is bubbled through water at 293 K how many millimoles of N dissolve in 1 litre of water Assume that N exerts a partial pressure of 0 987 bar Given that Henry s law constant for N at 293 K is 76 48 kbar NCERT Solved Example 097 har
3 26 H Partial pressure needed to dissolve 21 mg of CO in 100 gr of water at 298K is K for Co KPa m mol 2 3 a 14 KPa b 7 KPa c 121 KPa d 79 KPa ELITE SERIES for Sri Chaitanya Sr ICON
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3 26 H Partial pressure needed to dissolve 21 mg of CO in 100 gr of water at 298K is K for Co KPa m mol 2 3 a 14 KPa b 7 KPa c 121 KPa d 79 KPa ELITE SERIES for Sri Chaitanya Sr ICON
3 Consider the reaction Pb OH 4 2 aq Sn OH 4 2 aq Pb Sn OH 6 2 aq 2OH aq If equal moles of both Pb OH 4 2 and Sn OH 6 2 are initially present in its aqueous solution then after reaction 1 Vapour pressure of solution increases 2 Van t Hoff factor decreases 3 Boiling point of solution increases 4 Freezing poiont of solution increases
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3 Consider the reaction Pb OH 4 2 aq Sn OH 4 2 aq Pb Sn OH 6 2 aq 2OH aq If equal moles of both Pb OH 4 2 and Sn OH 6 2 are initially present in its aqueous solution then after reaction 1 Vapour pressure of solution increases 2 Van t Hoff factor decreases 3 Boiling point of solution increases 4 Freezing poiont of solution increases
that the solution is ideal III Show that if Raoult s law is applicable to one of the constituents of a binary liqu mixture at all composition it must be equally applicable to the other constituent
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that the solution is ideal III Show that if Raoult s law is applicable to one of the constituents of a binary liqu mixture at all composition it must be equally applicable to the other constituent
A 0 3 molal solution of a complex compound decreased the freezing point of pure water by 0 558 C The complex compound is known to show geometrical isomerism also which of the following may represent the complex compound Kf H O 1 86 K kg mol O CrCl3 NH3 3 O O Pt NH3 3 H2O 3 CI4 Pt NH3 2C14 Pt gly 2
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A 0 3 molal solution of a complex compound decreased the freezing point of pure water by 0 558 C The complex compound is known to show geometrical isomerism also which of the following may represent the complex compound Kf H O 1 86 K kg mol O CrCl3 NH3 3 O O Pt NH3 3 H2O 3 CI4 Pt NH3 2C14 Pt gly 2
b On dissolving 3 24 g of sulphur in 40 g of benzene boiling point of solution was higher than that of benzene by 0 81K K 2 53 K kg mol What is molecular formula of sulphur Atomic mass of S 32 g mol
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b On dissolving 3 24 g of sulphur in 40 g of benzene boiling point of solution was higher than that of benzene by 0 81K K 2 53 K kg mol What is molecular formula of sulphur Atomic mass of S 32 g mol
1 The vapour pressure of pure benzene and toluene at 40 C are 184 0 torr and 59 torr respectively Calculate the partial pressure of benzene and toluene the total vapour pressure of the solution mole fraction of benzene0 40 at solution and ass that the solution is ideal
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1 The vapour pressure of pure benzene and toluene at 40 C are 184 0 torr and 59 torr respectively Calculate the partial pressure of benzene and toluene the total vapour pressure of the solution mole fraction of benzene0 40 at solution and ass that the solution is ideal
36 ii Choose the correct option Mark only one oval 0000 In mixture A A Vmixing 0 and AH mixing 0 In mixture B A Vmixing 0 and AHmixing 0 In mixture A A Vmixing 0 and AH mixing 0 In mixture C A Vmixing 0 and AHmixing 0
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36 ii Choose the correct option Mark only one oval 0000 In mixture A A Vmixing 0 and AH mixing 0 In mixture B A Vmixing 0 and AHmixing 0 In mixture A A Vmixing 0 and AH mixing 0 In mixture C A Vmixing 0 and AHmixing 0
The Henry law constant for N and O2 gas dissolved in water at 298K are 6 x 109 N m and 2 5 109 N m A sample of water at 298K was equilibrium with air containing 20 mole oxygen and 80 mole nitrogen at 1 atm If the dissolved gases were separated from a sample of this water and then dried Then calculate the mole of oxygen gas in the dried gaseous sample
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The Henry law constant for N and O2 gas dissolved in water at 298K are 6 x 109 N m and 2 5 109 N m A sample of water at 298K was equilibrium with air containing 20 mole oxygen and 80 mole nitrogen at 1 atm If the dissolved gases were separated from a sample of this water and then dried Then calculate the mole of oxygen gas in the dried gaseous sample
Aqueous solution of glucose and compound X are place in beaker A and B respectively in closed chamber AA A contains 40 w W C6H12O6 aq Solution and B contains 10 aq w W Solution of X non electrolyte What is the molecular mass of compound X A B 30 40 50 35 Closed chamber
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Aqueous solution of glucose and compound X are place in beaker A and B respectively in closed chamber AA A contains 40 w W C6H12O6 aq Solution and B contains 10 aq w W Solution of X non electrolyte What is the molecular mass of compound X A B 30 40 50 35 Closed chamber
An amount of 1 mole of a non volatile solid is dissolved in 200 moles of water The solution is cooled to a temperature T K lower than the freezing point of solution to cause ice formation After removal of ice the remaining solution is heated to 373 K where the vapour pressure of solution is observed to be 740 mm Hg Select correct information Kf of water 2 K m 0 C 2 934 Kg of ice is formed at T K TK 273 2000 37 x 18 K Freezing point of the original solution is 10 18 C 1 Relative lowering of vapour pressure of the final solution is 38 4
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An amount of 1 mole of a non volatile solid is dissolved in 200 moles of water The solution is cooled to a temperature T K lower than the freezing point of solution to cause ice formation After removal of ice the remaining solution is heated to 373 K where the vapour pressure of solution is observed to be 740 mm Hg Select correct information Kf of water 2 K m 0 C 2 934 Kg of ice is formed at T K TK 273 2000 37 x 18 K Freezing point of the original solution is 10 18 C 1 Relative lowering of vapour pressure of the final solution is 38 4
5 What mass of non volatile solute urea needs to be dissolved in 150 g of water in order to decrease the vapour pressure of water by 35 What will be the molality of solution 1 23 88 3 25 47 2 31 12 4 29 92
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5 What mass of non volatile solute urea needs to be dissolved in 150 g of water in order to decrease the vapour pressure of water by 35 What will be the molality of solution 1 23 88 3 25 47 2 31 12 4 29 92
Definition of a mixture is anything that can b e seperated by a physical process accordin g to the video Q1 Is there anything that is made up of two or more substances but they cannot be sep erated by a physical process and so are clas sified as pure substances Q2 Is there anything which can be seperate d by a physical process but is classified as a pure substance
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Definition of a mixture is anything that can b e seperated by a physical process accordin g to the video Q1 Is there anything that is made up of two or more substances but they cannot be sep erated by a physical process and so are clas sified as pure substances Q2 Is there anything which can be seperate d by a physical process but is classified as a pure substance
0 Which of the given aqueous solution has the maximum boiling point K of H O 0 52 mol kg T H O 373 K 1 6 urea solution by weight of solution 2 18 glucose solution by weight of solution 3 34 2 sucrose solution by weight of solution 4 All have equal boiling points
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0 Which of the given aqueous solution has the maximum boiling point K of H O 0 52 mol kg T H O 373 K 1 6 urea solution by weight of solution 2 18 glucose solution by weight of solution 3 34 2 sucrose solution by weight of solution 4 All have equal boiling points
At 298 K the standard Gibbs energy of formation of H O is 256 5 kJ mol while standard Gibbs energy of its ionisation to H OH is 80 kJ mol What will be emf at 298 K of the cell volts H g 1 bar H IM OH IM O g 1 bar
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At 298 K the standard Gibbs energy of formation of H O is 256 5 kJ mol while standard Gibbs energy of its ionisation to H OH is 80 kJ mol What will be emf at 298 K of the cell volts H g 1 bar H IM OH IM O g 1 bar
3 A current of dry air is passed through a bulb containing 5 g of a solute in 100 g of water and then through water alone The losses in weight of the solution and pure water were respectively 0 78 g and 0 02 g Calculate relative lowering of vapour pressure 1 2 04 3 0 03 2 1 05 4 0 09
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3 A current of dry air is passed through a bulb containing 5 g of a solute in 100 g of water and then through water alone The losses in weight of the solution and pure water were respectively 0 78 g and 0 02 g Calculate relative lowering of vapour pressure 1 2 04 3 0 03 2 1 05 4 0 09
2 Molar solubility of helium nitrogen and oxygen are plotted against partial pressure of the gas at constant temperature Henry s law constant for these gases will lie in the following sequence 1 0 N He 3 O N He Solubility 0 Partial Pressure 2 O N He 4 ON H N He
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2 Molar solubility of helium nitrogen and oxygen are plotted against partial pressure of the gas at constant temperature Henry s law constant for these gases will lie in the following sequence 1 0 N He 3 O N He Solubility 0 Partial Pressure 2 O N He 4 ON H N He
B A current of dry air is passed through a bulb containing 5 g of a solute in 100 g of water and then through water alone The losses in weight of the solution and pure water were respectively 0 78 g and 0 02 g Calculate relative lowering of vapour pressure 1 2 04 3 0 03 2 1 05 4 0 09
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B A current of dry air is passed through a bulb containing 5 g of a solute in 100 g of water and then through water alone The losses in weight of the solution and pure water were respectively 0 78 g and 0 02 g Calculate relative lowering of vapour pressure 1 2 04 3 0 03 2 1 05 4 0 09
3 Solute A is a ternary electrolyte and solute B is non electrolyte If 0 1 M solution of solute B produces an osmotic pressure of 2P then 0 05 M solution of A at the same temperature will produce an osmotic pressure equal to 1 P 3 2P 2 1 5P 4 3P
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3 Solute A is a ternary electrolyte and solute B is non electrolyte If 0 1 M solution of solute B produces an osmotic pressure of 2P then 0 05 M solution of A at the same temperature will produce an osmotic pressure equal to 1 P 3 2P 2 1 5P 4 3P
55 Phenol associates in benzene to a certain extent to form a dimmer A solution containing 20 x 10 kg phenol in 1 kg of benzene has its freezing point depressed by 0 69 K The fraction of phenol is mm that has dimerised K for benzene 5 12 kg mol K
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55 Phenol associates in benzene to a certain extent to form a dimmer A solution containing 20 x 10 kg phenol in 1 kg of benzene has its freezing point depressed by 0 69 K The fraction of phenol is mm that has dimerised K for benzene 5 12 kg mol K
9 How many grams of sucrose M wt 342 should be dissolved in 100 g water in order to produce a solution with a 105 0 C difference between the boiling point and the freezing temperatures K 1 86 C m K 0 51 C m 1 34 2 g 3 342 g 2 72 g 4 460 g
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9 How many grams of sucrose M wt 342 should be dissolved in 100 g water in order to produce a solution with a 105 0 C difference between the boiling point and the freezing temperatures K 1 86 C m K 0 51 C m 1 34 2 g 3 342 g 2 72 g 4 460 g
12 The cryoscopic constant for acetic acid is 3 6 K kg mol A solution of 1 g of a hydrocarbon in 100 g of acetic acid freezes at 16 14 C instead of the usual 16 60 C The hydrocarbon contains 92 3 carbon What is the molecular formula 1 CH 3 CgH18 2 C6H12 4 C10H6
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12 The cryoscopic constant for acetic acid is 3 6 K kg mol A solution of 1 g of a hydrocarbon in 100 g of acetic acid freezes at 16 14 C instead of the usual 16 60 C The hydrocarbon contains 92 3 carbon What is the molecular formula 1 CH 3 CgH18 2 C6H12 4 C10H6
100 mL of FeSO4 solution required 200 mL of 0 1 M acidified KMnO4 solution for complete reaction What is the molarity of the FeSO4 solution MnO Fe H Mn Fe H O Select an answer A B 0 5 M 1 M 1 5 M
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100 mL of FeSO4 solution required 200 mL of 0 1 M acidified KMnO4 solution for complete reaction What is the molarity of the FeSO4 solution MnO Fe H Mn Fe H O Select an answer A B 0 5 M 1 M 1 5 M
4 Out of the following liquid pair which solution follows the positive deviation from Raoult s law 1 Acetone Chloroform 2 Water Nitric acid 3 Water Hydrochloric acid 4 Benzene Methanol
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4 Out of the following liquid pair which solution follows the positive deviation from Raoult s law 1 Acetone Chloroform 2 Water Nitric acid 3 Water Hydrochloric acid 4 Benzene Methanol
161 0 01 molal solution of Pt NH Cl in water had a freezing point depression of 0 054 C If K of water is 1 8 the correct formula for above molecule is A Pt NH3 4Cl CI B PI NH3 4Cl JCl C Pt NH3 4Cl CI D PI NH3 4 Cl4 A 168
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161 0 01 molal solution of Pt NH Cl in water had a freezing point depression of 0 054 C If K of water is 1 8 the correct formula for above molecule is A Pt NH3 4Cl CI B PI NH3 4Cl JCl C Pt NH3 4Cl CI D PI NH3 4 Cl4 A 168
A teacher wanted to give acid base titration to her students For that she prepared 1 HCI solution by dissolving 73 g of hydrochloric acid in one litre of water and ii sodium hydroxide solution by dissolving 0 46 g of sodium metal in one litre of water Find the volume of the hydrochloric acid solution required for complete neutralization of sodium hydroxide solution A 5 mL C 20 ml B 10 mL D 46 mL
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A teacher wanted to give acid base titration to her students For that she prepared 1 HCI solution by dissolving 73 g of hydrochloric acid in one litre of water and ii sodium hydroxide solution by dissolving 0 46 g of sodium metal in one litre of water Find the volume of the hydrochloric acid solution required for complete neutralization of sodium hydroxide solution A 5 mL C 20 ml B 10 mL D 46 mL
SolveLancer Test 4 4 g of polymer having carboxylic group is titrated with NaOH and requires 11 mL of 0 02 M NaOH Calculate average molar mass of Polymer g mol SolveLancer Test a 1 0 x 104 b 20 x 10 c 0 2 x 10 d 2 x 104
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SolveLancer Test 4 4 g of polymer having carboxylic group is titrated with NaOH and requires 11 mL of 0 02 M NaOH Calculate average molar mass of Polymer g mol SolveLancer Test a 1 0 x 104 b 20 x 10 c 0 2 x 10 d 2 x 104
Two liquids A and B form ideal solutions At 300 K the vapour pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm Hg At the same temperature if one more mole of B is added to this solution the vapour pressure of the solution increases by 10 mm Hg The vapour pressures of A and B in their pure states are respectively a PA 600 mm Hg and pg 400 mm Hg b p 550 mm Hg and pe 560 mm Hg c PA 450 mm Hg and pe 650 mm Hg d p 400 mm Hg and pg 600 mm Hg B
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Two liquids A and B form ideal solutions At 300 K the vapour pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm Hg At the same temperature if one more mole of B is added to this solution the vapour pressure of the solution increases by 10 mm Hg The vapour pressures of A and B in their pure states are respectively a PA 600 mm Hg and pg 400 mm Hg b p 550 mm Hg and pe 560 mm Hg c PA 450 mm Hg and pe 650 mm Hg d p 400 mm Hg and pg 600 mm Hg B
When acetone and chloroform are mixed together hydrogen bonds are formed between them Which of the following statements is correct about the solution made by mixing acetone and chloroform a On mixing acetone and chloroform will form an ideal solution b On mixing acetone and chloroform positive deviation is shown since the vapour pressure increases On mixing acetone and chloroform negative deviation is shown since there is decrease in vapour pressure d At a specific composition acetone and chloroform will form minimum boiling azeotrope
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When acetone and chloroform are mixed together hydrogen bonds are formed between them Which of the following statements is correct about the solution made by mixing acetone and chloroform a On mixing acetone and chloroform will form an ideal solution b On mixing acetone and chloroform positive deviation is shown since the vapour pressure increases On mixing acetone and chloroform negative deviation is shown since there is decrease in vapour pressure d At a specific composition acetone and chloroform will form minimum boiling azeotrope
1 what is the freezing point of a solution prepared by dissolving 1 00 mol of hf in 500 gram water is 3 8 c ii The Freezing point of a Solution of 1 00 mol Hcl In 500gr Water Is 7 4 c Calculate I In Both Cases And Point Out a Difference Between Hcl And HF Based On The Calculation
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1 what is the freezing point of a solution prepared by dissolving 1 00 mol of hf in 500 gram water is 3 8 c ii The Freezing point of a Solution of 1 00 mol Hcl In 500gr Water Is 7 4 c Calculate I In Both Cases And Point Out a Difference Between Hcl And HF Based On The Calculation
Which of the following helps in flocculation of lyophobic colloids O By adding opposite charge electrolyte O Electrophoresis Continuous dialysis O All of these
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Which of the following helps in flocculation of lyophobic colloids O By adding opposite charge electrolyte O Electrophoresis Continuous dialysis O All of these
Choose the correct option among the following O Lyophilic sols are reversible sols O Lyophilic sols are more stable than lyophobic sols O Lyophilic colloids are used to protect lyophobic colloids O All of these
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Choose the correct option among the following O Lyophilic sols are reversible sols O Lyophilic sols are more stable than lyophobic sols O Lyophilic colloids are used to protect lyophobic colloids O All of these
3 CaCl 4 1 w v BaCl 48 48 Molality of an aqueous solution of glucose Molar mass 180 gmol is 1 0 m 590 g of this solution is cooled and the temperature is kept at 3 C The amount of ice separated out is nearly K for water 2K kgmol is 1 333 g 2 500 g 3 167 g 4 200 g 3 1 w v CaCl 4 1 w v BaCl cita facena 1m fe fac a 590 gms f 1 333 g 2 500 g 3 167 g 4 200 g
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3 CaCl 4 1 w v BaCl 48 48 Molality of an aqueous solution of glucose Molar mass 180 gmol is 1 0 m 590 g of this solution is cooled and the temperature is kept at 3 C The amount of ice separated out is nearly K for water 2K kgmol is 1 333 g 2 500 g 3 167 g 4 200 g 3 1 w v CaCl 4 1 w v BaCl cita facena 1m fe fac a 590 gms f 1 333 g 2 500 g 3 167 g 4 200 g
1 5 points For an experiment on osmosis some dialysis tubing a semi permeable membrane s filled with 750 mL of a glucose C6H12O6 nonelectrolyte solution glucose cannot pass hrough the membrane If the reaction is carried out at 36 5 C and creates an osmotic pressure of 17 5 atm what would be the concentration of the glucose solution
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1 5 points For an experiment on osmosis some dialysis tubing a semi permeable membrane s filled with 750 mL of a glucose C6H12O6 nonelectrolyte solution glucose cannot pass hrough the membrane If the reaction is carried out at 36 5 C and creates an osmotic pressure of 17 5 atm what would be the concentration of the glucose solution
x 1 0 where p is the vapour pressure of pure component I at the same Similarly for component 2 P P x 11 where p represents the vapour pressure of the pure compone temperature solution phase in the container will be the sum of the partial pre According to Dalton s law of partial pressures the total pressure components of the solution and is given as P P P Substituting the values of p and p we get Pa X P X P 1 x p x P P P 9 x Following conclusions can be drawn from equation 2 16 Total vapour pressure over the solution can be related to the mole Ini component Total vapour pressure over the solution varies linearly with the mole fran component 2 or 1 H Depending on the vapour pressures of the pure components 1 and 2 U pressure over the solution either decreases or increases with the increas mole fraction of either component 1 or 2 a solution gives a linear plot as shown in Fig 2 A plot of p or p versus the mole fractions lines I and II also pass through the points for s and x are equal to unity Similarly the plot Ptotal versus x is also linear Fig 2 3 The minu of Ptotal is p and the maximum value is that component 1 is less volatile than compone P 20 PP B P P Mole fraction III II x 0 X 1 P P The composition of vapour phase in equilibrium solution is determined by the partial pressures components If y and y are the mole fractions components 1 and 2 respectively in the vapour then using Dalton s law of partial pressures P Y Ptotal P2 Y2 Ptotal In general P Y Ptotal 2 3 plot of vapour pressure and mole fraction ideal solution at constant temperature ashed lines I and II represent the partial are of the components It can be seen the plot that p and p are directly tional to x and x respectively The 12
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x 1 0 where p is the vapour pressure of pure component I at the same Similarly for component 2 P P x 11 where p represents the vapour pressure of the pure compone temperature solution phase in the container will be the sum of the partial pre According to Dalton s law of partial pressures the total pressure components of the solution and is given as P P P Substituting the values of p and p we get Pa X P X P 1 x p x P P P 9 x Following conclusions can be drawn from equation 2 16 Total vapour pressure over the solution can be related to the mole Ini component Total vapour pressure over the solution varies linearly with the mole fran component 2 or 1 H Depending on the vapour pressures of the pure components 1 and 2 U pressure over the solution either decreases or increases with the increas mole fraction of either component 1 or 2 a solution gives a linear plot as shown in Fig 2 A plot of p or p versus the mole fractions lines I and II also pass through the points for s and x are equal to unity Similarly the plot Ptotal versus x is also linear Fig 2 3 The minu of Ptotal is p and the maximum value is that component 1 is less volatile than compone P 20 PP B P P Mole fraction III II x 0 X 1 P P The composition of vapour phase in equilibrium solution is determined by the partial pressures components If y and y are the mole fractions components 1 and 2 respectively in the vapour then using Dalton s law of partial pressures P Y Ptotal P2 Y2 Ptotal In general P Y Ptotal 2 3 plot of vapour pressure and mole fraction ideal solution at constant temperature ashed lines I and II represent the partial are of the components It can be seen the plot that p and p are directly tional to x and x respectively The 12
Solution A B C and D have ph values equal t o 12 7 4 6 5 respectively 1 Which solution will evolved with ammoni a with ammonia chloride Why 2 Which solution will evolved with co3 with Na2co3 Why 3 What type of reaction occurs between sol ution a and and c
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Solution A B C and D have ph values equal t o 12 7 4 6 5 respectively 1 Which solution will evolved with ammoni a with ammonia chloride Why 2 Which solution will evolved with co3 with Na2co3 Why 3 What type of reaction occurs between sol ution a and and c
2 Among the following solution mixtures the one with highest boiling point is A 100ml 0 1M acetic acid and 100ml 0 1M NaCl B 100ml 0 1M CaCl2 and 100ml 0 1M NaCl C 100ml 0 1 M Na2SO4 and 100ml 0 1M BaCl2 D 100ml 0 1 M sucrose and 100ml 0 1M NaCl
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2 Among the following solution mixtures the one with highest boiling point is A 100ml 0 1M acetic acid and 100ml 0 1M NaCl B 100ml 0 1M CaCl2 and 100ml 0 1M NaCl C 100ml 0 1 M Na2SO4 and 100ml 0 1M BaCl2 D 100ml 0 1 M sucrose and 100ml 0 1M NaCl
2 Colligative properties depend on O The nature of the solute particles dissolved in solution O The number of solute particles in solution O The physical properties of the solute particles dissolved in solution O The nature of solvent particles
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2 Colligative properties depend on O The nature of the solute particles dissolved in solution O The number of solute particles in solution O The physical properties of the solute particles dissolved in solution O The nature of solvent particles
4 Mixture of volatile components A and B has total vapour pressure in torr P 254 119 A total where XA is the mole fraction of A in mixture Hence p and pare in torr 1 254 119 3 135 254 2 119 254 4 154 119
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4 Mixture of volatile components A and B has total vapour pressure in torr P 254 119 A total where XA is the mole fraction of A in mixture Hence p and pare in torr 1 254 119 3 135 254 2 119 254 4 154 119
ing point of an aqueous solution of benzoic acid is 101 C A solution of equal molality was prepared by dissolving benzoic acid in zene Then identify correct statements from the following ven that for water K 0 83k kg mol K 1 86k kg mol and for benzenek 2 0k kg mol and boiling point is 75 C Freezing point of the aqueous solution of benzoic acid is 2 24 C Boiling point of the solution of benzoic acid in benzene is 77 4 C Observed molar mass of benzoic acid in water is equal to its value in presence of benzene as the solvent Observed molar mass of benzoic acid in water is less than to its value in presence of benzene as the solvent
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ing point of an aqueous solution of benzoic acid is 101 C A solution of equal molality was prepared by dissolving benzoic acid in zene Then identify correct statements from the following ven that for water K 0 83k kg mol K 1 86k kg mol and for benzenek 2 0k kg mol and boiling point is 75 C Freezing point of the aqueous solution of benzoic acid is 2 24 C Boiling point of the solution of benzoic acid in benzene is 77 4 C Observed molar mass of benzoic acid in water is equal to its value in presence of benzene as the solvent Observed molar mass of benzoic acid in water is less than to its value in presence of benzene as the solvent