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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1401. |
`30mL` of `CH_(3)OH (d = 0.780 g cm^(-3))` and `70 mL` of `H_(2)O (d = 0.9984 g cm^(-3))` are mixed at `25^(@)C` to form a solution of density `0.9575 g cm^(-3)`. Calculate the freezing point of the solution. `K_(f)(H_(2)O)` is `1.86 kg mol^(-1)K`. Also calculate its molarity. |
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Answer» Correct Answer - `-19.91^(@)C, 7.63M` `CH_(3) OH rarr V_(1) = 30 mL, d_(1) = 0.798 g//mL` `m_(1) = 23.94 g` `H_(2)O rarr V_(2) = 70 mL, d_(2) = 0.9984 g//mL` `m_(2) = 69.888g` `m_(T) = 93.828g` `d_("solution") = 0.9575 g//mL` `V_("solution") = 98 mL` `DeltaT_(f) = (1.86 xx 23.94 xx 1000)/(32xx 69.888) = 19.91` `T_(f) =- 19.91^(@)C` `M = (23.94)/(32 xx 0.98) = 7.63M` |
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| 1402. |
At `25^(@)C, 1 mol` of A having a vapour pressure of 100 torr and 1 mole of B having a vapour pressure of 300 torr were mixed. The vapour at equilibrium is removed, condensed and the condensate is heated back to `25^(@)C`. The vapour now formed are again removed, recondensed and analyzed. what is the mole fraction of A in this condensate? |
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Answer» Correct Answer - `0.1` `P_(A)^(@) = 100, P_(B)^(@) = 300, X_(A) = X_(B) = (1)/(2)` `P_(T) = 200` `Y_(A) = (100 xx(1)/(2))/(200) =(1)/(4)` On condensation `X_(A) = (1)/(4),X_(B) = (3)/(4)` `P_(T) = 100 xx (1)/(4) +300 xx (3)/(4) = 250` `Y_(A) = (25)/(250) = 0.1` on further condensation `X_(A) = 0.1` |
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| 1403. |
If `V_(1)mL` of `C_(1)` solution `+V_(2)mL` of `C_(2)` solution are mixed together then calculate final concentration of solution and final osmotic pressure. If initial osmotic pressure of two solutions are `pi_(1)` and `pi_(2)` respectively? |
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Answer» `(C_(1)V_(1)+C_(2)V_(2))/(V_(1)+V_(2))` `pi_(1)=C_(1)RT, C_(1)=((pi_(1))/(RT)) , pi_(2)=C_(2)RT,C_(2)=((pi_(2))/(RT))` `pi=((C_(1)V_(1)+C_(2)V_(2))/(V_(1)+V_(2)))RT` `pi=((pi_(1)V_(1)+pi_(2)V_(2))/(V_(1)+V_(2)))` |
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| 1404. |
Differentiate between molarity and molality. What is the effect of change in temperature of a solution on molarity and molality ? |
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Answer» Molarity is defined as the ratio no. Of moles of solute to the volume of solution in litres. Molality is defined as the no. Of moles of solute in 1 Kg of solvent. On changing temp.molarity changes as volume changes but molality remains unaffected as mass of solvent remains unchanged on changing temp. |
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| 1405. |
The concentration of pollutant in `ppm (w//w)`. That has been measured at `450mg` per `150kg` of sample isA. `3 ppm`B. `6ppm`C. `3000ppm`D. `330ppm` |
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Answer» Correct Answer - A Given, In `150xx10^(3)gm` of sample `450mg` is present so in `10^(6)gm (450)xx10^(-3))/(150xx10^(3))xx10^(6)` `3ppm` |
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| 1406. |
Select correct statement ?A. Heats orf vaporisation for a pure solvent and for a solution are similar because similar intermolecular forces between solvent molecules must be overcome in both case.B. Entropy change between solution and vapour is smaller than the entropy change between pure solvent and vapourC. Boiling point of the solution is larger than that of the pure solventD. All are correct statements |
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Answer» Correct Answer - D All are facts. We should remember that, Entropy of solution is more than entropy of pure solvent. So the difference in entropy will be less in case of solution. |
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| 1407. |
(a) Differentiate between molarity and molality for a solution. How does a change in temperature influence their values ? (b) Calculate the freezing point of an aqueous solution containing `10.50g` of `MgBr_(2)` in `200g` of water. (Molar mass of `MgBr_(2)=184g`) (`K_(f)` for waer `=1.86K kg "mol"^(-1)`) |
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Answer» Correct Answer - (a) Molality: It is the number of the solute dissolve per kilogram `(kg)` of the solvent. It is denoted by `therefore` Molality `(m)` =("Moles of solute")/("Mass of solvent in"kg)` `tehrefore m=(w_(2)xx1000)/(M_(2)xxw_(1))` where `w_(1)=` mass of solvent, `w_(2)=` mass of solute and `M_(2)=` Molar mass of solute. Molarity: It is the number of moles of the solute dissoved per litre of the solutions. It is denoted by `M`. `therefore` Molarity `(M)`=("Moles of solute")/("Volume of solution in litre")` `therefore M=(w_(2)xx1000)/(M_(2)xxV)` where `w_(2)=` mass of solute, `M_(2)=` molar mass of solute and `V=` Volume of solution. The molarity and molality of a solution will be nearly same if the mass of solvent is nerarly equal to the volume of solution. Molarity decreases because volume of the solution increases with increases in temperature. But molaity remains unaffected. (b) Molecular mass of `MgBr_(2)=184g` mass of `MgBr_(2)`=184g)` `DeltaT_(f)=(K_(f)xxW_(B)xx1000)/(M_(B)xxW_(A))=(1.86xx10.50xx1000)/(184xx200)=0.530` Freezing point of solution `=0-530=-0.530^(@)C. |
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| 1408. |
Predict the osmotic pressure order for the following: (I) `0.1 N` urea (II) `0.1 N NaCl` (III) `0.1 N Na_(2)SO_(4)` (IV) `0.1 N Na_(3)PO_(4)` |
| Answer» Correct Answer - (a) Urea `ltNaClltNa_(2)SO_(4)ltNa_(3)PO_(4)` (b) `6.15` atm | |
| 1409. |
Assertion : Ethyl slcohol and water form maximum boiling azetrope. Reason : Attractive forces in solution tend to increase.A. If both assertion are reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are corrct but reason is not correct explanation for assertion.C. If assertion is corrct but reason is incorrect.D. If assertion and reason both are incrrect. |
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Answer» Correct Answer - d Correct Assertion. Ethyl alcohol and water from minimum boiling azetrope. Correct Reason. Attractive forces in solution tend to decrease. |
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| 1410. |
What is meant by abnormal molecular mass of solute? Discuss the factors which bring abnormallty in experimentally determined molecular masses of solutes using colligative properties. |
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Answer» Correct Answer - The molecular mass obtained with the help of colligative property sometimes is different from normal molecular mass, it is called abnormal molecular mass. The factors which bring abnormallity are: Association. When the solute particles undergo association, number of particles become less and molecular mass determined with the help of colligative property will be more. Dissociation. It leads to increase in number of particles, therefore, increase in colligative property, therefore, decrease in molecular weight because colligative property in inversely proportional to molecular weight |
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| 1411. |
Assertion Istonic solution donot show any osmosis when placed side by side. Reson : Isotnic solutions have same solute concentration.A. If both assertion are reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are corrct but reason is not correct explanation for assertion.C. If assertion is corrct but reason is incorrect.D. If assertion and reason both are incrrect. |
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Answer» Correct Answer - c Correct Reason. Isotonic solution have same molar concentration of the solute. |
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| 1412. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution `M` is prepared by mixing ethanol and water. Thus moel fraction of ethanol in the mixture is `0.9`. Given: Freezing point depression constant of water `(K_(f)^("water")) = 1.86 K kg mol^(-1)` Freezing point depression constant of ethanol `(K_(f)^("ethanol")) = 2.0 K kg mol^(-1)` Boiling point elevation constant of water `(K_(b)^("water")) = 0.52 K kg mol^(-1)` Boiling point elevation constant of ethanol `(K_(b)^("ethanol")) = 1.2 K kg mol^(-1)` Standard freezing point of water `= 273 K` Standard freezing point of ethanol `= 155.7K` Standard boiling point of water `= 373 K` Standard boiling point of ethanol `= 351.5 K` vapour pressure of pure water `= 32.8 mm Hg` Vapour pressure of pure ethanol `= 40 mm Hg` Molecualr weight of water `= 18 g mol^(-1)` Molecular weight of ethanol `= 46 g mol^(-1)` In asweering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of the solution `M` is :A. `268.7 K`B. `268.5 K`C. `234.2 K`D. `150.9 K` |
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Answer» Correct Answer - D Solution `M` is a mixture of ethanol and water, where ethanol is the solvent because the mole fraction of ethanol is `0.9`. The mole fraction of water =`0.1` Molarity of `H_(2)O(m_(H2O))=(chi_(2) xx 1000)/(chi_(1) xx Mw_(1))` `=(0.1 xx 1000)/(0.9 xx 46) =2.415` `DeltaT_(f)=K_(f) xx m =2 xx 2.415 =4.83` Freezing point =`155.7-4.83 =150.87 K` |
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| 1413. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution `M` is prepared by mixing ethanol and water. Thus moel fraction of ethanol in the mixture is `0.9`. Given: Freezing point depression constant of water `(K_(f)^("water")) = 1.86 K kg mol^(-1)` Freezing point depression constant of ethanol `(K_(f)^("ethanol")) = 2.0 K kg mol^(-1)` Boiling point elevation constant of water `(K_(b)^("water")) = 0.52 K kg mol^(-1)` Boiling point elevation constant of ethanol `(K_(b)^("ethanol")) = 1.2 K kg mol^(-1)` Standard freezing point of water `= 273 K` Standard freezing point of ethanol `= 155.7K` Standard boiling point of water `= 373 K` Standard boiling point of ethanol `= 351.5 K` vapour pressure of pure water `= 32.8 mm Hg` Vapour pressure of pure ethanol `= 40 mm Hg` Molecualr weight of water `= 18 g mol^(-1)` Molecular weight of ethanol `= 46 g mol^(-1)` In asweering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution `M` is:A. `39.3 mm Hg`B. `36.0 mm Hg`C. `29.5 mm Hg`D. `28.8 mm Hg` |
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Answer» Correct Answer - B According to Raoult,s law `P=P_(A)^(@)chi_(A) =40 xx 0.9 =36 mm Hg` In the paragraph, it has been directed to take solute as non-volatile, thus `H_(2)O` does not contribute in total vapour pressure. |
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| 1414. |
A solution `M` is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is `0.9` Given: Freezing point depression constant of water `(K_(f)^(water)=1.86 K kg mol^(-1))` Freezing point depression constant to ethanol `(K_(f)^(ethanol))=2.0 K kg mol^(-1))` Boiling point elevation constant of water `(K_(b)^(water))=0.52 K kg mol^(-1))` Boiling point elevation constant of ethanol `(K_(b)^(ethanol))=1.2 K kg mol^(-1))` Standard freezing point of water = `273 K ` Standard freezing point of ethanol = `155.7 K ` Standard boiling point of water = `373 K ` Standard boiling point of ethanol = `351.5 K ` Vapour pressure of pure water =`32.8 mm Hg` Vapour pressure of pure ethanol =`40 mm Hg` Molecular weight of water =`18 g mol^(-1)` Molecular weight of ethanol =`46 g mol^(-1)` In anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution `M` isA. `39.3 mm Hg`B. `36.0 mm Hg`C. `29.5 mm Hg`D. `28.8 mm Hg` |
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Answer» Correct Answer - B `P_("Total")=P_(A)^(@) = P_(A)^(@)chi_(A) [:.` Solute is to be taken as non-volatile] `=40 xx 0.9 =36 mm Hg` |
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| 1415. |
A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have equalA. Potential energyB. Total energyC. Kinetic energyD. Intermolecular forces |
| Answer» In liquid `hArr` vapour equilibrium, molecule have same kinetic energy. | |
| 1416. |
A solution `M` is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is `0.9` Given: Freezing point depression constant of water `(K_(f)^(water)=1.86 K kg mol^(-1))` Freezing point depression constant to ethanol `(K_(f)^(ethanol))=2.0 K kg mol^(-1))` Boiling point elevation constant of water `(K_(b)^(water))=0.52 K kg mol^(-1))` Boiling point elevation constant of ethanol `(K_(b)^(ethanol))=1.2 K kg mol^(-1))` Standard freezing point of water = `273 K ` Standard freezing point of ethanol = `155.7 K ` Standard boiling point of water = `373 K ` Standard boiling point of ethanol = `351.5 K ` Vapour pressure of pure water =`32.8 mm Hg` Vapour pressure of pure ethanol =`40 mm Hg` Molecular weight of water =`18 g mol^(-1)` Molecular weight of ethanol =`46 g mol^(-1)` In anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution `M` such that the mole fraction of water in the solution becomes `0.9`. The boiling point of this solution isA. `380.4 K`B. `376.2 K`C. `375.5 K`D. `354.7 K` |
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Answer» Correct Answer - B `chi_(H2O)=0.9 (solvent)`,`chi_C_(2)H_(5)OH=0.1 (solute)` Molality of `C_(2)H_(5)OH (m_(C_(2)H_(5)OH))=(chi_(2) xx 1000)/(chi_1 xx Mw_(1))` `(0.1 xx 1000)/(0.9 xx 18)` `DeltaT_(b)=K_(b) xx m` `=0.52 xx (0.1 xx 1000)/(0.9 xx 18) =3.2 K` Boiling point =`373 + 3.2 =376.2 K` |
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| 1417. |
A liquid is in equilibrium with its vapour at its boiling point . On an average the molecules in the two phases have equal :A. intermolecular forcesB. potential energyC. temperatureD. None of these |
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Answer» Correct Answer - D At the point of equilibrium the rate of two opposing processes is same. |
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| 1418. |
A solution showing a large positive deviation from ideal behaviour haveA. lower boiling point than both the componentsB. higher boiling point than both the componentsC. `DeltaH_("mixing")` is positiveD. `DeltaH_("mixing")` is negative |
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Answer» Correct Answer - A::C A solution of two volatile liquids showing a large positive deviation from ideal behaviour have `DeltaH_("mixing")` negative and lower boiling point than both the components. |
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| 1419. |
In the depression of freezing point experiment, it is found that the:A. The vapour pressure of the solution is less than that of pure waterB. The vapour pressure of the solution is more than that of pure solventC. Only solute molecules solidify at the freezing pointD. Only solvent molecules solidify at the freezing point |
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Answer» Correct Answer - A::D In the depression of freezing point experiment, vapour pressure of the solution is less than that of pure solvent and only the solvent molecules solidify at the freezing point. |
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| 1420. |
Osmosis is involved in one or more processes. (I) Interchange of nutrients and waste products between tissue cells and their surroundings. (II)Reverse osmosis (III)Excrection of urine (IV) Evaporation Select the correct processes.A. `(I) and (III)`B. `(I),(II) and (III)`C. `(I),(II) and (IV)`D. All of these |
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Answer» Correct Answer - B (I) interchange of nutrients and waste products between tisssue cells andtheir surroundings.These are the processes involving osmosis. (II) Reverse opmosis is used in desalination of sea water. (III)Excretion of urine. |
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| 1421. |
In reverse osmosis the solvent molecules flow from …………to solvent side. |
| Answer» Correct Answer - D | |
| 1422. |
A solution of two liquids A and B showing a large negative deviation from ideal behaviour haveA. higher boilint point than both the componentsB. `DeltaH_("mixing")` is negativeC. `DeltaV_("mixing")` is negativeD. A.........B interactions are stronger than A........A and B........B interactions. |
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Answer» Correct Answer - A::B::C::D A solution of two volatile liquids showing a large negative deviation have higher boiling point than both the compounts, `DeltaH_("mixing")` is negative, `DeltaV_("mixing")` is negative, A…..B interactions are stronger than A….A and B….B interaction. |
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| 1423. |
Which of the following is correct for a non-ideal solution of liquids A and B showing negative deviation?A. `DeltaH_(mix) =- ve`B. `DeltaV_(mix) =- ve`C. `DeltaS_(mix) =+ve`D. `DeltaS_(mix) =- ve` |
| Answer» Correct Answer - A::B::D | |
| 1424. |
CCI4 and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. |
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Answer» CCI4 is a non-polar covalent compound. Water is a polar compound. CCI4 can neither form H– bonds with water molecules nor can it break H- bonds between water molecules. Therefore, it is insoluble in water. Ethanol is a polar compound and can form H- bonds with water, which is a polar solvent, therefore it is miscible with water in all proportions. |
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| 1425. |
How is it that alcohol and water are miscible inall proportions ? |
| Answer» Both alcohol and water are polar in nature. Moreover, these are also involved in the intermolecular hydrogen bonding. Therefore, these are miscible in all proportions. | |
| 1426. |
`CCl_(4)` and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. |
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Answer» `C Cl_(4)` is a non-polar covalent compound, whereas water is a polar compound.`C Cl_(4)` can neither form H-bonds with water molecules nor can it break H-bonds in water molecules. Therefore, it is insoluble in water. Ethanol is a polar compound and can form H-bonds with water, which is a polar solvent. Therefore,it is miscible with water in all proportions. |
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| 1427. |
Discuss biological and industrial importance of osmosis. |
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Answer» Hint: The process of osmosis is of immense biological and industrial importance as is evident from the following examples : (i) Movement of water from soil into plant roots and subsequently into upper portion of the plant is partly due to osmosis. (ii) Preservation of meat against bacterial action by adding salt. (iii) Preservation of fruits against bacterial action by adding sugar. Bacterium in canned fruit loses water through the process of osmosis, shrivels and dies. (iv) Reverse osmosis is used for desalination of water. |
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| 1428. |
Which of the following is not an industrial or biological importance of osmosis ?A. Movement of water from soil into plant roots and upper portion of plant .B. Salting of meat to prevent bacterial actionC. Reverse osmosis for desalination of sea waterD. Filling of ink in a fountain pen |
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Answer» Correct Answer - D It is capillary action. |
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| 1429. |
Discuss biological and industrial applications of osmosis. |
| Answer» For answer, consult section 12. | |
| 1430. |
In the phenomenon fo osmosis, the membrance allow passage of :A. Solute onlyB. Solvent onlyC. Soth solute and solventD. None of these |
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Answer» Correct Answer - B Solute molecules are not small enough to pass through pores of semipermeable membrance. |
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| 1431. |
An aqueous solution at `-2.55^(@)C`. What is its boiling point `(K_(b)^(H_(2)O)=0.52 K m^(-1)`,`K_(f)^(H_(2)O)=1.86 K m^(-1)`?A. `107.0^(@)C`B. `100.6^(@)C`C. `100.1^(@)C`D. `100.7^(@)C` |
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Answer» Correct Answer - D `DeltaT_(f)=2.55^(@)C = K_(f).m :. M=2.55/K_(f)` `DeltaT_(b)=K_(b)m rArr DeltaT_(b)` `=2.55 xx K_(b)/K_(f)=2.55 xx 0.52/1.86^(@)C)=0.7^(@)C` `rArr T_(b) = 100 + 0.7 =100.7^(@)C` |
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| 1432. |
The relative decrease in `VP` of an aqueous glucose dilute solution is found to be `0.018`. Hence, the elevation in boiling point is (it is given `1 molal` aqueous urea solution boils at `100.54^@)C` at `1 atm` pressure)A. `0.018^(@)`B. `0.18^(@)`C. `0.54^(@)`D. `0.03^(@)` |
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Answer» Correct Answer - C `chi_(B)=(DeltaP)/P_(A)^(@)=0.018` `DeltaT_(b)=K_(b).m rArr 0.54=K_(b) = xx1=K_(b)=0.54` `m=chi_(B)/(1-chi_(B))xx1000/(Mw_(A))` `m=0.018/0.982xx1000/18=1.0 rArr DeltaT_(b) = K_(b) xx m=0.54^(@)C` |
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| 1433. |
`2.5 g` of a substance is present in `200 mL` of solution showing the osmotic pressure of `60 cm Hg` at `15^(@)C`. Calculate the molecular weight of substance.What will be the osmotic pressure if temperature is raised to `25^(@)C`? |
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Answer» Given that, `W_(2)=2.5 g` ,`V=200/1000 L` `pi=60/70` atm, `T=288 K` `:. piV=W_(2)/(Mw_2)RT` `60/70 xx 200/1000 = 2.5/(Mw_2)xx 0.0821 xx288 rArr Mw_(2)=374.38` Also, `pi_(1)/pi_(2)=T_(1)/T_(2)` `60/pi_(2)=288/298 rArr pi_(2) = 62.08 cm` |
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| 1434. |
A `10 g` mixture of glucose and urea present in `250 mL` solution shows the osomotic pressure of `7.4 atm` at `27^(@)C`. Calculate `%` composition of mixture. |
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Answer» Let the mixture contains a g glucose and b g urea. `:.a+b=10` …(i) Also, `7.4 xx 250/1000=[a/180 + b/60]xx0.0821xx300` `3a + b =13.52` …(ii) Solving Eqs. (i) and (ii), we get `a=1.76 g` `b=8.24 g` `%` of glucose =`(1.76/10) xx 100 =17.6` `%` of urea =`(8.24/10) xx 100 =82.4` |
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| 1435. |
For a dilute solution containing 2.5g of a non-volatile non-electrolyte solute in 100g of water, the elevation in boiling point at 1 atm pressure is `2^(@)"C"`. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take `K_(b)`=0.76 K kg `"mol"^(-1)`)A. 724B. 740C. 736D. 718 |
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Answer» Correct Answer - A `DeltaT=K_(b)xxm` `2=0.76xxm` `m=(2)/(0.76)=2.63` Mole fraction of solute can be calculated as `m=(x_(B)xx1000)/((1-x_(B))m_(A))` `2.63=(x_(B)xx1000)/((1-x_(B))xx18)` `x_(B)=0.045` `(p_(0)-p)/(p_(0))=x_(B)` `(760-p)/(760)=0.045` or `p=725.8` mm It is close to 724 mm] |
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| 1436. |
The vapour pressure of a solution of 5g of non-electrolyte in 100g of water at a particular temperature is `2985 N m^(-2)`. The vapour pressure of pure water at that temperature is `3000 N m^(-2)`. The molecular weight of the solute isA. `180`B. `90`C. `270`D. `200` |
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Answer» Correct Answer - a `(P_(A)^(@)-P_(S))/P_(S)=W_(B)/M_(B)xxM_(A)/W_(A)` `(3000-2985)/(2985)=((5g))/((M_(B)))xx((18g))/((100g))` `M_(B)=180 (by calculation)` |
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| 1437. |
The vapour pressure of a solution of 5g of non-electrolyte in 100g of water at a particular temperature is `2985 N m^(-2)`. The vapour pressure of pure water at that temperature is `3000 N m^(-2)`. The molecular weight of the solute isA. 180B. 90C. 270D. 200 |
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Answer» Correct Answer - A `((3000-2985)Nm^(-2))/(3000Nm^(-2))=(5//M_(B))/(100//18)` (For a dilute solution) `(15xx100)/(18xx3000)=(5)/(M_(B))or M_(B)=180` |
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| 1438. |
Vapour pressure of a solution of `5g`of non-electrolyte in `100g`water at a particular temperature is`2985N//m^(2)`.The vapour pressure of pure water is`3000N//m^(2)`.The molecular weight of the solute isA. `60`B. `120`C. `180`D. `380` |
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Answer» Correct Answer - C `(P^(0)-P^(s))/(P^(0))=((W_(2))/(M_(2)))/((W_(1))/(M_(1)))=(3000-2985)/(3000)=((5)/(M_(2)))/((100)/(18)) or M_(2)=180` |
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| 1439. |
At `100^(@)C` the vapour pressure of a solution of `6.5g` of an solute in `100g` water is `732mm`.If `K_(b)=0.52`, the boiling point of this solution will be :A. `103^(@)C`B. `101^(@)C`C. `100^(@)C`D. `102^(@)C` |
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Answer» Correct Answer - B At.`B.P.P_(0)=760 "torrr"` for elevation of B.P. `(P_(0)-P_(s))/(P_(s))=(W_(B)//M_(B))/(W_(A)//M_(A))` `(760-732)/(732)=(6.5//M)/(100//18)` On solving `M=32` `=1xx0.52xx((6.5)/(32))/((100))xx1000=1` `DeltaT_(B)=T_(B)^(S)-T_(B)^(0)soT_(B)^(S)=DeltaT_(B)+T_(B)^(0)` `T_(B)^(S)=1+100=101^(@)C` |
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| 1440. |
Vapour pressure of a solution of `5g`of non-electrolyte in `100g`water at a particular temperature is`2985N//m^(2)`.The vapour pressure of pure water is`3000N//m^(2)`.The molecular weight of the solute isA. 60. 0B. 120. 0C. 180. 0D. 380. 0 |
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Answer» Correct Answer - C `(DeltaP)/P^(@)=chi_(B)= rArr (DeltaP)/P^(@)=chi_(B)=(5//Mw_(B))/(5//Mw_(B) + 100//18)` `rArr Mw_(B)=179.28 g "mol"^(-1)` |
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| 1441. |
`5g` of compound `A` was dissolved in `100g` of water at `303K`. The vapour pressure of the solution was found to be `4.16` kilopascal. If the vapour pressure of pure water is `4.24 kPa` at this temperature, what is the molecular mass of `A`? |
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Answer» Given Mass of solute `(A)=5g` Mass of solvent (water)`=100g` vapour pressure of the solution`=4.16` killopasical vapour pressure of solvent `=4.24kPa` Asked molecular mass of solute `=?` Formulae: `(P_(A)^(0)-P_(A))/(P_(A)^(0))=x_(B)=(W_(B)/(M_(B)))/(W_(A)/M_(A)+W_(B)/(M_(B)))` Explanation : `P_(A)^(0)=` vapour pressure of pure solvent `P_(A)=` vapour pressure of solution `X_(B)=` Mole fraction of solute `W_(B)=` Molecular weight solute `W_(A)=` Mass of solvent `M_(A)=` Molecular weight of solvent Substitution `&` Calculation `(4.21-4.16)/(4.24)=(5)/(M_(B)/(100/(18)+(5)/(M_(B))))rArr(0.08)/(4.23)=(5)/(M_(B)/(100/(18)+(5)/(M_(B))))rArr(0.08xxM_(B))/(M_(B))+0.08xx(100)/(M_(B))xx4.24` `rArr(0.40)/(M_(B))+(8)/(18)=(21.20)/(M_(B))rArr(20.80)/(M_(B))=(8)/(18)` `rArr M_(B)=(20.80xx18)/(8)=46.8g mol^(-1)`. |
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| 1442. |
Diffine the following mofes of expressing the concentration of a solution? Which of which of these modes are independent of temperature and why ? `{:((a),w//w("mass percentage"),(b),V//V("volume percentage")),((c),w//V("mass by volume percentage"),(d),"ppm" ("part per million")),((e),X("mole fraction"),(f),M("molarity")),((g),m("molality"),,):}` |
| Answer» For answer, consult Section 2 (Eexpression Concentration of solution) | |
| 1443. |
Why is vapour pressure of solvent lowered on adding non-volatile solute to it ? |
| Answer» Surface area available for evaporation of the solvent gets reduced. | |
| 1444. |
The lowering of vapour pressure of the solvent takes place on dissolving a non-volatile solute becauseA. the density of the solution increasesB. the surface tension of the solution decreasesC. the molecules of the solvent on the surface are replaced by the molecules of the soluteD. the mole fraction of solvent is less than 1 |
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Answer» Correct Answer - D Vapour pressure does not depend upon surface area of a volatile liquid `(p)/(p^(@)) = x_("solvent")` In case of a solution `x_("solvent") lt 1`. `:. p lt p^(@)`. |
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| 1445. |
If `P^(@)` the vapour pressure of a pure solvent and P is the vapour pressure of the solution prepared by dissolving a non volatile solute in it. The mole fraction of the solvent `X_(A)` is given by:A. `(P^(@) -P)/(P^(@)) = X_(A)`B. `(P^(@)-P)/(P) =X_(A)`C. `(P)/(P^(@)) = X_(A)`D. `P^(@) - P = X_(A)` |
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Answer» Correct Answer - B `P = X_(A) P^(@)`: |
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| 1446. |
Define the freezing point. Why the freezing point of a solvent gets lowered on dissolving a non-volatile solute into it? |
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Answer» Freezing point of a solution is defined as the temperature at which the vapour pressure of its liquid phase becomes equal to the vapour pressure of its solid phase.when a non volatile solute is added to a volatile solvent its vapour pressure decreases and it would become equal to that of solid solvent at lower temperature.Hence its freezing point gets depressed. |
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| 1447. |
10 gram of solute with molecular mass 100 gram `mol^(-1)` is dissolved in 100 gram solvent to show `0.3^(@)C` elevation in boiling point. The value of molal ebuilioscopic constant will be:A. 10B. 3C. `0.3`D. un predictable |
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Answer» Correct Answer - C `0.3 = ((1)/(10))/(100) xx 1000 xx k_(b)` `k_(b) = 0.3` |
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| 1448. |
When a liquid that is immiscible with water was steam distilled at `952^(@)C` at a total pressure of `748` torr, the distillate contained `1.25g` of the liquid per gram of water. The vapour pressure of water is `648` torr at `95.2^(@)C`. What is the molar mass of liquid?A. `7.975 g//"mol"`B. `166 g//"mol"`C. `145.8g//"mol"`D. None of these |
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Answer» Correct Answer - A::B::C::D For two immischible liquid , `P_(A)^(@)=P_("total")-P_(H_(2^(O)))^(@)=748-648rArr100` `(W_(A))/(W_(B))=(P_(A)^(@)M_(A))/(P_(B)^(@)M_(B)) , M_(A)=(1.25)/(1)xx(648xx18)/(100)rArr145.8` |
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| 1449. |
The freezing point of one molal `NaCl`solution assuming `NaCl`to be `100%`dissociated in water is (molal depression constant=`1.86)`A. `-1.86^(@)C`B. `-3.72^(@)C`C. `+1.8^(@)C`D. `+3.72^(@)C` |
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Answer» Correct Answer - B For `NaCl i=2` `DeltaT_(f)=2K_(f)m=2xx1.86xx1=3.72` `T_(s)=T=DeltaT_(f)=0.3.72=-72^(@)C` |
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| 1450. |
The freezing point of `1` molal `NaCl` solution assuming `NaCl` to be `100%` dissociated in water is:A. `-1.86^(@)C`B. `-3.72^(@)C`C. `+1.86^(@)C`D. `+3.72^(@)C` |
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Answer» Correct Answer - B `DeltaT_(f)=1.86 xx 1 xx 2 = 3.72`, `:.T_(f)=0-3.72 =-3.72^(@)C` `NaCl` dissociates to give experimental molality `=1 xx 2 =2` |
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