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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.
| 301. |
A solution of a non-volatile solute in water freezes at `-0.30^(@)C`. The vapour pressure of pure water at `298 K` is `23.51 mm Hg` and `K_(f)` for water is `1.86 degree//molal`. Calculate the vapour pressure of this solution at `298 K`. |
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Answer» Correct Answer - `P_(s)=23.442 mm Hg` We know that `DeltaT_(f)=K_(f) xx "Molality of solution"` From given data, `0.3=1.86xx"Molality of solution"` Hence, molality of solution =`0.3/1.86=0.161 m` Lowering in vapour pressure (for dilute solution) `(P^(@)-P_(S))/P^(@)=(W_(2)xxMw_(1))/(Mw_(1) xx W_(1))` =`(W_(2)/((Mw_(2))xxW_(1))xx(Mw_(1))/1000)/1000` `rArr P^(@)-P_(S))/P^(@)` =`"Molality of solution" xx ("Molecular weight of solvent")/1000` or `23.51-P_(S)/23.51=0.61 xx 18/1000` By solving this equation, we get Vapour pressure of solution `(P_(S))=23.44 mm` |
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| 302. |
The vapour pressure of acetone at `20^(@)C` is 185 torr. When `1.2 g` of non-volatile substance was dissolved in `100g` of acetone at `20^(@)C` its vapour pressure was 183 torr. The moalr mass `(g mol^(-1))` of the substance is:A. 128B. 488C. 32D. 64 |
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Answer» Correct Answer - D |
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| 303. |
Gold numbers of protective colloids A, B, C and D are `0.50, 0.01, 0.10` and `0.005` respectively. The correct order of their protective powers isA. `C lt B lt D lt A`B. `A lt C lt B lt D`C. `B lt D lt A lt C`D. `D lt A lt C lt B` |
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Answer» Correct Answer - B Higher the goal number, lesser will be the protective power of colloid. |
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| 304. |
If `M_("normal")` is the normal molecular mass and `alpha` is the degree of ionization of `K_(3)[Fe(CN)_(6)]`, then the abnormal molecular mass of the complex in the solution will be :A. `M_("normal")`B. `M_("normal")(1+3alpha)^(-1)`C. `M_("normal")(1+alpha)^(-1)`D. equal to `M_("normal")` |
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Answer» Correct Answer - B `{:(,K_(3)[Fe(CN)_(6)],hArr,3K^(+),+,[Fe(CN)_(6)]^(3-)),(At t=0,1,,0,,0),("After ionization",(1-alpha),,3alpha,,alpha,i=1+3alpha):}` Since, `i=(M("normal"))/(M_("normal"))` `therefore (1+3alpha)/(1)=(M_("normal"))/(M_("abnormal"))` |
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| 305. |
Define the following terms: (i) Abnormal molar mass (ii) van't Hoff factor (i) |
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Answer» (i) If the molar mass calculated by using any of the colligative properties to be different than theoretically expected molar mass. (ii) Extent of dissociation or association or ratio of the observed colligative property to calculated colligative property. Detailed Answer: (i) When the molar mass calculated with the help of colligative property, is different from theoretical molar mass, it is called as abnormal molar mass. (ii) The ratio of the observed (experimental) value of a colligative property to the normal (calculated) value of the same property is called as van't Hoff factor (i). |
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| 306. |
Class 12 Chemistry MCQ Questions of Solutions with Answers? |
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Answer» MCQ Questions for class 12 chemistry of Solutions with Answers help you specialize in fundamentals to assist students with basic concepts. This may cause you to understand the solutions thoroughly thereby assist you in scoring high marks within the Class 12 board examination and graduate entrance examinations. The answers that are provided Questions help the students in analyzing if their answers are correct. These MCQ Questions aid students in revising their theoretical concepts. Learn Important Class 12 Chemistry MCQ Questions of Solutions with Answers at Sarthaks eConnect includes chapter-wise solutions for Chemistry. These Class 12 Chemistry MCQ Questions are prepared by the experts to assist the students of class 12 strength their foundation in Chemistry. Practice MCQ Question for Class 12 Chemistry chapter-wise 1. The atmospheric pollution is generally measured in the units of (a) mass percentage 2. 234.2 gm of sugar syrup contains 34.2 gm of sugar. What is the molal concentration of the solution (a) 0.1 3. Atomic mass is equal to (a) number of electrons of an atom 4. The law which indicates the relationship between solubility of a gas in liquid and pressure is (a) Raoult’s law 5. Among the following substance the lowest vapour pressure is exerted by (a) water 6. Partial pressure of a solution component is directly proportional to its mole fraction. This is known as (a) Henry’s law 7. Which of the following solutions shows positive deviation from Raoult’s law? (a) Acetone + Aniline 8. The system that forms maximum boiling azeotrope is (a) Acetone-chloroform 9. A plant cell shrinks when it is kept in a (a) hypotonic solution 10. The relative lowering in vapour pressure is proportional to the ratio of number of (a) solute molecules to solvent molecules 11. The osmotic pressure of a solution can be increased by (a) increasing the volume 12. Sprinkling of salt helps in clearing the snow covered roads in hills. The phenomenon involved in the process is (a) lowering in vapour pressure of snow 13. For carrying reverse osmosis for desalination of water the material used for making semipermeable membrane is (a) potassium nitrate 14. What will be the degree of dissociation of 0.1 M Mg(NO3)2 solution if van’t Hoff factor is 2.74? (a) 75% 15. Which of the following units is useful in relating concentration of solution with its vapour pressure? (a) Mole fraction 16. At equilibrium the rate of dissociation of a solid solute in a volatile liquid solvent is (a) less than the rate of crystallisation 17. If molality of the dilute solution is doubled, the value of molal depression constant (Kf) will be? (a) doubled 18. Temperature dependent term among the following is - (a) Molality 19. An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase? (a) Addition of NaCl 20. A solution of acetone in ethanol is- (a) Shows a negative deviation from Raoult's law 21. The mixture that forms maximum boiling azeotrope is- (a) Heptane + Octane 22. Isotonic solutions have same- (a) Vapour pressure 23. Among the following options, 0.1 M aqueous solution will have the lowest freezing point is- (a) Potassium Sulphate 24. A two phase system consisting of a liquid dispersed in another liquid is known as- (a) Foam 25. The van't Hoff factor [i] for a dilute aqueous solution of the strong electrolyte barium hydroxide is? (a) 0 Answer : 1. Answer : (d) ppm Explanation: Since the amount of pollutants present in the atmosphere is very small, atmospheric pollution is measured in parts per million (ppm). 2.Answer : (b) 0.5 Explanation: Weight of water in sugar syrup = 234.2 − 34.2 = 200gm Molecular weight of sugar = 342 Molal concentration \(=\frac{34.2}{342}\times\frac{1000}{200}\) Molal concentration = 0.5 3. Answer : (b) sum of the numbers of electrons and protons of an atom Explanation: The atomic mass of an atom is an empirically measured property, which is equivalent to the sum mass of protons, neutrons, and electrons that make up the atom (with a small adjustment for nuclear binding energy). 4. Answer : (b) Henry’s law Explanation: Henry's Law is the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. 5. Answer : (d) mercury Explanation: Mercury exerts the lower vapour pressure because of the strong bond among its atom. 6. Answer : (b) Raoult’s law Explanation: The relationship is known as the Raoult's law which states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. 7. Answer : (b) Acetone + Ethanol Explanation: Acetone + ethanol is an example of solutions showing positive deviation from Raoult's law. 8. Answer : (a) Acetone-chloroform Explanation: acetone – chloroform is a non-ideal solution and has a negative deviation. It is a maximum boiling azeotrope. 9. Answer : (b) hypertonic solution Explanation: Hypertonic solution has high osmotic pressure. When a plant cell is placed in hypertonic solution water will diffuse out of the cell resulting in shrinking of the cell. 10.Answer : (c) solute molecules to the total number of molecules in solution Explanation: Relative lowering of vapour pressure. The vapour pressure of a liquid is the pressure of the vapour which is in equilibrium with that liquid. The vapour pressure of a solvent is lowered when a non-volatile solute is dissolved in it to form a solution. 11. Answer : (b) increasing the number of solute molecules Explanation: The osmotic pressure of a solution increases as the number of solute molecules increases. Excess pressure produced on the solution side owing to osmosis is known as osmotic pressure. The osmotic pressure of a solution increases as the number of solute molecules increases. 12. Answer : (b) depression in freezing point of snow Explanation: When salt is spread over snow covered roads, snow starts melting from the surface because depression of freezing point of water takes place due to addition of salt. It helps in clearing of roads. Hence, the phenomena is depression in freezing point which helps in clearing the snow covered roads in hilly areas. 13.Answer : (c) cellulose acetate Explanation: Cellulose acetate, potassium ferrocyanide, etc. are used as semipermeable membrane for carrying out reverse osmosis. 14. Answer : (b) 87% Explanation: \(Mg(NO_3)_2\longrightarrow\) \(Mg^{2+}+2NO^-_3\) \(\alpha=\frac{i-1}{n-1}\) \(=\frac{2.74-1}{3-1}\) \(=\frac{1.74}{2}\) = 0.87 Degree of Dissociation = 0.87(100) = 87% 15. Answer : (a) Mole fraction Explanation: Mole fraction is useful in relating vapour pressure with concentration of solution. According to Raoult's law, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. 16. Answer : (c) equal to the rate of crystallisation Explanation: This means that the rate of crystallisation is equal to the rate of the dissolution for a solid solute in a volatile liquid. At equilibrium the rate of dissolution will be equal to the rate of crystallisation. 17. Answer : (d) unchanged Explanation: If the molality of a dilute solution is doubled, the value of the molal depression constant (Kf) remains unchanged. Because molal depression is constantly dependent on the nature of the solvent, not on solute. 18. Answer : (b) Molarity Explanation: molarity is the concentration term which is temperature dependent. 19. Answer : (d) Addition of water Explanation: However, the addition of solvent, ie, dilution increases the surface area of the liquid surface, thus results in increased vapour pressure. Hence, the addition of water to the aqueous solution of (1 molal) KI, result in increased vapour pressure. 20. Answer : (b) Shows a positive deviation from Raoult's law Explanation: A solution of acetone in ethanol shows a positive deviation from Raoult's law due to miscibility of these two liquids with a difference of polarity and length of the hydrocarbon chain. 21. Answer : (b) Water + Nitric acid Explanation: The mixture that forms maximum boiling azeotrope is water and nitric acid. This mixture of water and nitric acid shows negative deviation from Roult's law and it will form maximum boiling azeotrope. 22. Answer : (c) Osmotic pressure Explanation: An isotonic solution is a solution that has the same osmolarity or solute concentration as another solution. 23. Answer : (a) Potassium Sulphate Explanation: Urea and glucose do not dissociate in solution. Sodium chloride gives two ions and potassium sulphate gives three ions per formula unit. Therefore, the effective number of particles is maximum in potassium sulphate. Hence, its freezing point is lowest. 24. Answer : (c) Emulsion Explanation: An emulsion is a mixture of two or more liquids that are normally immiscible (unmixable or unblendable) owing to liquid-liquid phase separation. Emulsions are part of a more general class of two-phase systems of matter called colloids. 25. Answer : (d) 3 Explanation: Ba(OH)2 is a strong electrolyte, so its 100% dissociation occurs in solution. Van't off factor(i) = Total number of ions present in the solution. \(Ba(OH)_2\longrightarrow \) \(Ba^{2+}+2OH^-\) So, the total number of ions in the solution is 3. Click here to practice MCQ Question for Solution Class 12 |
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| 307. |
Two liquids `HNO_3` (A) and water (B) form a maximum boiling azeotrope when mixed in the ratio of 68% and 32% respectively . It meansA. A-B interactions are stronger than A -A and B-B interactionsB. A-B interactions are weaker than A -A and B-B interactionsC. vapour pressure of solution is more than the pure componentsD. vapour pressure of solution is less since only one component vaporises . |
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Answer» Correct Answer - A Since the new interactions (A-B) are stronger than (A-A) and (B-B) interactions , the solution shows negative deviation, i.e. less vapour pressure and hence higher boiling point . |
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| 308. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solution molecules ae added to get homogeneous solution. These are called colligative properties. Application colligative properties are very useful in day-to-day life. One of its example 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. The mole fraction of ethanol in the mixture is `0.9` Given: Freezing point depression of water `(K_(f)^("water"))=1.86K kg "mol"^(-1)` Freezing point depression constant of ethanol `(K_(f)^("ethanol")=2.0K kg "mol"^(-1)` Boiling point elevation constant of water `(K_(b)^("water")=0.52K kg "mol"^(-1)` Boiling point elevation constant of ethanol `(K_(b)^("ethanol")=1.2K kg "mol"^(-1)` Standard freezing point of water`=273K` Standard freezing point of ethanol`=155.7K` Standard boiling point of water `=373K` Stadndard boiling point of ethanol `=351.5K` Vapour pressure of pure water `=32.8mm Hg ` Vapour pressure of pure ethanol`=40mm Hg` Molecular weight of water `=18 g "mol"^(-1)` Molecular weight of ethanol `=46g "mol"^(-1)` In aswering the following questions, consider the solution to be ideal dilute solutions and solutes to be non volatile and non-dissociative. The freezing point (in `.^(@)C`) of a solution containing `0.1g` of `K_(3)[Fe(CN)_(6)]` (Mol. Wt. `329`) in `100g` of water `(K_(f)=1.86K kg "mol"^(-1)` is :A. `-2.3xx10^(-2)`B. `-5.7xx10^(-5)`C. `-5.7xx10^(-3)`D. `-1.2xx10^(-2)` |
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Answer» Correct Answer - A `DeltaT_(f)=ixxK_(f)xxm` `=4xx1.86xx(0.1)/(329xx0.1)=2.3xx10^(-2)` `rArrT_(f)=0-2.3xx10^(-2)=-2.3xx10^(-2)^(@)C`. |
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| 309. |
Twenty grams of a solute are added to `100g` of water at `25^(@)C`. The vapour pressure of pure water is `23.76 mmHg`, the vapour pressure of the solution is `22.41` Torr. (a) Calculate the molar mass of the solute. (b) What mass of this solute is required in `100g` of water of reduce the vapour pressure ot one-half the value for pure water? |
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Answer» Correct Answer - (a) `60 g//"mol"`, (b) `333.6g` (a) Relative lowering of vapour pressure `=` mole fraction of solute `(23.76-22.41)/(23.76)=(20//M)/(20//M+100//18)=0.0568` so `M=60 g//"mol"` (b) Let `w` be the mass of solute reqd. `& P=(P^(@))/(2)` so `(P^(@)-P)/(P^(@))=(w//60)/(w//60+10//18)` or `(P^(@)-P^(@)//2)/(P^(@))=(w//60)/(w//60+5.56)` `therefore w=333.6gm` |
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| 310. |
`10ml` of sulphuric acid solution `(sp.gr.=1.84)` contains `98%` weight of pure acid. Calculate the volume of `2.5 M NaOH` solution required to just neutralise the acid. |
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Answer» Wt of solute `=10xx1.84xx(98)/(100)gm` So moles of solute `=(18.4)/(98)xx(98)/(100)=0.184` `n_(H+)=2xx0.184` `2xx0.184=(2.5xxV)/(1000)` `V=147.2` |
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| 311. |
A sample of `H_(2)SO_(4)` (density `1.8 g mL^(-1)`) is labelled as `74.66%` by weight. What is molarity of acid? (Give answer in rounded digits) |
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Answer» Let `V=1l t=1000ml` mass of solution `=1800gm` So mass of solute `=1800xx(74.66)/(100)gm` So molarity `=` moles of solute in `1l t=(18xx74.66)/(98)=13.71M` |
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| 312. |
We have 100 mL of 0.1 MKCI solution . To make it 0.2 MA. evaporate `50mL` waterB. evaporate `50mL` solutionC. add `0.1` mol `KCl`D. add `0.01` mol `KCl` |
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Answer» Correct Answer - A::B::D Initial moles of `KCl=(100)/(1000)xx0.1=0.01` Let `x` moles of `KCl` are added, so `0.2=(0.001+x)/(100//1000)` or `x=0.01` |
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| 313. |
If urea (aq) solution at `500K` has `O.P.=2.05` atm. `&` glucose solution at `300K` has `OP=1.23` atm. If `200mL` of `I^(st)` solution `& 400 ml` of `2^(nd)` solution are mixed at `400K` then calculate `O.P.` of resulting solution at `400K` (assume molarity is not dependent on temp.) |
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Answer» `C_("urea")=(2.05)/(Rxx500)=0.05 rArr V_("urea")=2CO mL` `C_("glucose")=(1.23)/(Rxx300)=0.05 rArr V_("glucose")=400mL` `C_("total")=(C_(1)V_(1)+C_(2)V_(2))/(V_(1)+V_(2))=(0.05xx200+0.05xx400)/(600)=0.05` `pi=CRT=0.05xx0.082xx400` `pi=1.64` atm |
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| 314. |
If osmotic pressure of `1M` aqueous solution of `H_(2)SO_(4)` at `500K` is `90.2` atm. Calculate `K_(a2)` of `H_(2)SO_(4)`. Give your answer after multiplying `1000` with`K_(a2)`. (Assuming ideal solution). (Given: `K_(a1)` of `H_(2)SO_(4)` is `oo,R=0.082 l t-atm//mol-K)`. |
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Answer» Correct Answer - `300` `{:(H_(2)SO_(4),rarr,H^(+),+,HSO_(4)^(-),,,HSO_(4)^(-)hArr,H^(+),+,SO_(4)^(2-)),(1,,,,,,1,1,,),(-,,1,,1,,1-x,1+x,,x):}` Total concentration, `C_(T)=(1+X)+(1-X)+X=2+X`. `pi=C_(T)RT rArr 90.2=(2+X)xx0.082xx500 rArr x=0.2`. `therefore [H^(+)]=1.2M, [SO_(4)^(2-)]=0.2M, [HSO_(4)^(-)]=1-0.2=0.8`. `therefore Ka_(2)=([H^(+)][SO_(4)^(2-)])/([HSO_(4)^(-)])=(1.2xx0.2)/(0.8)=0.3` `1000 K_(a2)=1000xx0.3=300` |
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| 315. |
Persons are medically considered to have lead poisoning if they have a concentration greater than `10` micrograms of lead per decilitre of blood. What is the concentration in parts per billion?A. `1000`B. `100`C. `10`D. `1` |
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Answer» Correct Answer - B `10mug` per decilitre `rArr10xx10^(-6)g` in `100 mL` `therefore 10^(9)` parts (`1` billion)has `=(10xx10^(-6))/(100)xx10^(9)g=100g` |
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| 316. |
If the density of lake water is 1.25 g `mL^(-1)` and it contains 92 g of `Na^(+)` ions per kg of water, calculate the molality of `Na^(+)` ions in the lake. |
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Answer» `"Molality of "Na^(+)"inos"(m)=("No. of moles of " Na^(+)"ions")/("Mass of water in kg")` `=((92g)//(23g mol^(-1)))/((1 kg))=4" mol kg"^(-1)=4m` |
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| 317. |
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n*octane and explain. Cyclohexane, KCl, CH3OH, CH3CN. |
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Answer» (i) Cyclohexane and n-octane both are non-polar. Hence they mix completely in all proportions. (ii) KCl is an ionic compound while n-octane is nonpolar. Hence, KCl will not dissolve at all in n-octane. (iii) CH3OH and CH3CN both are polar but CH3CN is less polar than CH3OH. As the solvent is non-polar, CH3CN will dissolve more than CH3OH is n-octane. Thus the order of solubility will be KCl< CH3OH < CH3CN < Cyclohexane. |
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| 318. |
Suggest the most important type of intermolecular attractive interaction in the following pairs.(i) n-hexane and n-octane (ii) I2 and CCl4 (iii) NaClO4 and water (iv) methanol and acetone (v) acetonitrile (CH3CN) and acetone (C3H6O). |
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Answer» (i) Both are non-polar. Hence, inter molecular interactions in them will be London/ dispersion forces. (ii) Both are non-polar. Hence, inter molecular interactions in them will be London/ dispersion forces. (iii) NaClO4 gives Na+ and ClO4 - ions in the solution while water is polar molecule. Hence, intermolecular interactions in them will be ion – dipole interactions. (iv) Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions. (v) Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions. |
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| 319. |
The vapour pressure of a pure liquid A is 60 torr at 300K. It forms an ideal solution with another liquid B. The mole fraction of B is 0.25 and the total pressure of the solution is 82.5 torr at 300K. The vapour pressure of pure liquid B at 300K isA. 37.5B. 150C. 75D. 187.5 |
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Answer» Correct Answer - B `82.5 = 60 xx 0.75 +x xx 0.5`or `x = 150` torr |
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| 320. |
Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?A. 20B. 50C. 100D. 200 |
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Answer» Correct Answer - A `(p^@-p_A)/p_A^@=n_B/n_A` `(2-1)/2=W_B/M_BxxM_A/W_A rArr 1/2=1/M_Bxx200/20` `rArr M_B=(200xx2)/20`=20 |
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| 321. |
If a `6.84%`(wt,vol.)solutionof cane sugar(mol.Wt.=`342`)is isotonic with `1.52%(wt.//vol.)`solution of thiocarbamide ,then the molecular weight of thiocarbamide isA. `152`B. `76`C. `60`D. `180` |
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Answer» Correct Answer - B `C_(1)=C_(2)` `(w_(1))/(M_(1)xxV_(1))=(w_(2))/(M_(2)xxV_(2))rArr (6.84)/(342xx100)=(1.52)/(M_(2)xx100)M_(2)=76` |
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| 322. |
The freezing point of ether was lowered by `0.60^(@)C`on dissolving `2.0g` of phenol in `100g` of ether. Calculate the molar mass of phenol and comment on the result. Given: `K_(f)` (ether)` = 5.12 K kg mol^(-1)`. |
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Answer» Correct Answer - `170.7 g mol^(-1)` `DeltaT_(f) = (1000 K_(f)W_(2))/(M_(2)W_(1))` `M_(2) = (1000 xx 5.12 xx 2)/(0.6 xx 100) M_(2) = 170.6 gm//mol` |
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| 323. |
What is the freezing point of a solution contains `10.0g`of glucose`C_(6)H_(12)O_(6)`, in `100g` of `H_(2)O` ? `K_(f)=1.86^(@)C//m`A. `-0.186^(@)C`B. `+0.186^(@)C`C. `-0.10^(@)C`D. `-1.03^(@)C` |
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Answer» Correct Answer - D `DeltaT_(f)=K_(F)xxmrArr1.86xx(10xx1000)/(180xx100)=1.03^(@)C` `rArr T=-1.03^(@)C` |
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| 324. |
4.0g of substance A dissolved in 100g `H_(2)O` depressed the freezing point of water by `0.1^(@)C` while 4.0g of another substance B depressed the freezing point by `0.2^(@)C`. Which one has higher molecular mass and what is the relation ? |
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Answer» Correct Answer - `M_(A)=2M_(B)` |
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| 325. |
What should be the osmotic pressure of a solution of urea in water at `30^(@)C` which has boiling point 0.052 K higher than pure water ? Assume molarity and molality to be the same. `K_(b)` for water is `0.52 "Kg kg mol"^(-1)` |
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Answer» Correct Answer - 2.487 atm |
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| 326. |
Explain the solubility of gases in liquids. |
Answer»
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| 327. |
The solubility of gases in liquids:A. increases with increase in pressure and temperatureB. decreases with increase in pressure and temperatureC. increases with increases in pressure and decrease in temperatureD. decreases with increases in pressure and increase in temperature |
| Answer» Correct Answer - C | |
| 328. |
Which of the following will form non-ideal solution?A. `C_(2)H_(5)OH` and waterB. `HNO_(3)` and waterC. `CHCl_(3)` and `CH_(3)COCH_(3)`D. `C_(6)H_(6)` and `C_(6)H_(5)CH_(3)` |
| Answer» Correct Answer - A::B::C | |
| 329. |
Given at `350Kp_(A)^@=300 "torr"` and `p_(B)^@=800 "torr"` the composition of the mixture having a normal boiling point of `350K` isA. `X_(A)=0.08`B. `X_(A)=0.06`C. `X_(A)=0.04`D. `X_(A)=0.02` |
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Answer» Correct Answer - A `760=300X_(A)+800(1-X_(A))` `rArr760=800-500X_(A)` `rArr 500X_(A)=40` `:.X_(A)=(40)/(500)=0.08` |
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| 330. |
In mixture A and B ,components show -ve deviations as:A. `DeltaV_("mix")` is +veB. A-B interation is weaker than A-A and B-B interationC. `DeltaH_("mix")` is +veD. A-B interation is stronger than A-A and B-B interation |
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Answer» Correct Answer - D It is required condition along with `Delta_(V)=-ve and Delta_(H)=-ve` |
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| 331. |
The vapour pressure of the solution of two liquids A`(P^@=80mm)`and B`(P^@=120mm)`is found to `100mm` when `x_(A)=0.4` .The result shows thatA. solution exhibits ideal behaviourB. solution shows positive deviationsC. solution shows negative deviationsD. solution will show postive deviations for lower concentration and negative deviations for higher concentrations. |
| Answer» Correct Answer - C | |
| 332. |
The vapour pressure of the solution of two liquids A`(P^@=80mm)`and B`(P^@=120mm)`is found to `100mm` when `x_(A)=0.4` .The result shows thatA. solution exhibits ideal behaviourB. solution shows positive deviationsC. solution shows negative deviationsD. solution will show positive deviations for lower concentration and negative deviations for higher concentrations. |
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Answer» Correct Answer - C `P_("total")=0.4xx80+0.6xx120` `=104 gt 100` mm of Hg. |
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| 333. |
What mass of sugar `C_(12)H_(22)O_(11)(M_(0)=342)` must be dissolved in `4.0 kg` of `H_(2)O` to yield a solution that will freeze at `-3.72^(@)C`. (Take `K_(f)=1.86^(@)C m^(-1)`) |
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Answer» `DeltaT_(f)=3.72^(@)C Mw_("sugar") =342` `DeltaT_(f)=K_(f)[(W_(B)/(Mw_(B))xx1000)/W_(A)]` `rArr W_(B) = (Delta_(f)T W_(A)Mw_(B))/(1000 K_(f))` `rArr W_(B) = (3.72xx4xx10_(3)xx342)/(1000xx1.86) =2736 g` |
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| 334. |
How much ethly alcohol much be added ot 1.0 litre of water so that the solution may treeze at `-10^(@)C. (K_(f) "for water =1.88 K kg" mol^(-1), "density of water=1 g "mL^(-1)).` |
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Answer» Correct Answer - 247.3g Step I. Calculation of molality of the solution. `DeltaT_(f)=K_(f)xxm` `DeltaT_(f)=0^(@)C-(-10^(@)C)=10^(@)C=10 K, K_(f)=1.86" K kg mol"^(-1)` `m=(DeltaT_(f))/K_(f)=((10K))/((1.86" K kg mol"^(-1)))=5.376" mol kg"^(-1)=5.376 m.` Step II. Calculation of amount of ethyl alcohol to be added. `"Molality (m)"=("Mass of ethyl alcohol / Molar mass")/("Mass of water / 1000")` `5.376" mol kg"^(-1)=(W_(B)//(46"g mol"^(-1)))/(1" kg")` `W_(B)=(5.376" mol kg"^(-1))xx(46" g mol"^(-1))xx(1 kg)=247.3 g.` |
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| 335. |
Calculate the freezing point depression and boiling point elevation of a solution of `10.0 g` of urea `(M_(B)=60)` in `50.0 g` of water at `1 atm`. pressure.`K_(b)` and `K_(f)` for water `0.52^(@)C m^(-1)` and `1.86^(@)C m^(-1)` respectively. |
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Answer» Depression in freezing point =`DeltaT_(f)` and elevation in boiling point =`DeltaT_(b)` `DeltaT_(f)=K_(f)((W_(B)/(Mw_(B)))/W_(A)xx1000)=1.86(((10)/(60))/(50)xx1000)=6.2^(@)C` `DeltaT_(b)=K_(b)((W_(B)/(Mw_(B)))/W_(A)xx1000)=0.52(((10)/(60))/(50)xx1000)=1.7^(@)C` Freezing point `(T_(f))=T_(f)^(@)-DeltaT_(f) =0-6.2 =-6.2^(@)C` Boiling point `(T_(b))=T_(b)^(@)-DeltaT_(b) =100 + 1.7 =-101.7^(@)C` |
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| 336. |
The elevation in boiling point of a solution of non-electrolyte in `C CI_(4)` is 0.60. What is the depression in freezing point for the same solution ? `K_(f)(C CI_(4))=30"K kg " mol^(-1), K_(b)(C CI_(4))=5.02 "K kg" mol^(1)`. |
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Answer» Correct Answer - 3.59 K `DeltaT_(f)=K_(f)xxm" and " DeltaT_(b)=K_(b)m ` `(DeltaT_(f))/(DeltaT_(b))=K_(f)/K_(f),DeltaT_(b)=((30"Kg mol"^(-1)))/((5.02"K kg mool"^(1)))xx(0.60K)=3.59 K.` `DeltaT_(f)=K_(f)/K_(b)xxDeltaT_(b)=((30" K J mol"^(-1)))/((5.02" K kg mol"^(-1)))xx(0.60 K)=3.59 K.` |
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| 337. |
In liquid `C CI_(4)`, which of the following has maximum solubility?A. `CI_(2)`B. `Br_(2)`C. `I_(2)`D. `KCI` |
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Answer» Correct Answer - C Out of the given choices, `I_(2)` has maximum solubility in `C CI_(4)`. |
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| 338. |
Which of the following is not a characteristic property of the polar liquids?A. They have high boiling pointsB. They have high heat of evaporationC. They have low viscosityD. They have low vapour pressure |
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Answer» Correct Answer - C Polar liquids have gigh viscosity due to strong dipole-dipole interaction. |
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| 339. |
Ebullioscopic constant is the boiling point elevation when the concentration of solution is (a) 1 m (b) 1 M (c) 1 mass% (d) 1 mole fraction of solute |
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Answer» (a) 1 m Ebullioscopic constant is the boiling point elevation when the concentration of solution is 1 m. |
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| 340. |
Calculating osmotic pressure: The fomula for low molecular mass starch is `(C_(6)H_(10)O_(5))_(n)`, where n averages `2.00xx10^(2)`. When `0.798g` of starch is dissolved in `100.0 mL` of water solution, what is the osmotic pressure at `25^(@)C`? Strategy: Work out the molecular mass of `(C_(6)H_(10)O_(5))_(200)`, and use it to obtain the molarity of the starch solution. Substitute into the formula for the osmotic pressure, `pi=CRT`. |
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Answer» Step 1. Calculate the molarity The molecular mass of `(C_(6)H_(10)O_(5))_(200)` is `200 [6xx12+10xx1+5xx16]=32,400 u`. Therefore, the number of moles in `0.798 g` of starch is `n("moles")=(mass_("starch"))/(molar mass_("starch"))=(0.798 g "starch")/(32.400 g " starch"//mol " starch")` `=0.0000246 mol` starch `=2.46xx10^(-5)` mol starch The molarity of the solution is `M=n_(starch)/V_(mL)xx(1000 mL)/L` `=(2.46xx10^(-5) mol)/(100.0 mL)xx(1000 mL)/L` `=2.46xx10^(-4) mol L^(-1)` solution Step 2. Calculate osmotic pressure The osmotic pressure at `25^(@)C` is `pi=CRT` `=(2.46xx10^(-4) mol L^(-1))(0.082 (L atm)/(K mol))(298 K)` `=6.02xx10^(-3)` atm `=6.02xx10^(-3) "atm" xx(760 mmHg)/(1 "atm")` `=4.58 mmHg` |
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| 341. |
In calculating osmotic pressure the concentration of solute is expressed in (a) molarity (b) molality (c) mole fraction (d) mass per cent |
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Answer» (a) molarity In calculating osmotic pressure the concentration of solute is expressed in molarity. |
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| 342. |
Blood is purified by :A. DialysisB. Electro-osmosisC. CoagulaionD. Filtration |
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Answer» Correct Answer - A Blood is colloidal solution which can be purified by dialysis. |
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| 343. |
Name the substances which are used by deep sea divers to neutralize the toxic effects of nitrogen dissolved in the blood. |
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Answer» Helium (11%) |
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| 344. |
Given below are few mixtures formed by mixing two components . Which of the following binary mixtures will have same composition in liquid and vapour phase ? (i)Ethanol + Chloroform (ii)Nitric acid + Water (iii) Benzene + Toluene (iv)Ethyl chloride + Ethyl bromideA. (i) and (iii)B. (i) and (ii)C. (i), (ii) and (iii)D. (iii) and (iv) |
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Answer» Correct Answer - B (iii) and (iv) will form ideal solutions hence do not form azeotropes. Azeotropes have same composition in liquid and vapour form when distilled |
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| 345. |
...... a non – aqueous solution. |
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Answer» Br2 in CCl4 a non – aqueous solution. |
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| 346. |
… is an example for gaseous solution. |
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Answer» Camphor in nitrogen gas is an example for gaseous solution. |
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| 347. |
Match the column I with column II.column I column IIA. 95 g(i) CaSO4 .2H2OB. Henry's Law(ii) V/VC. Volume Percentage(iii) Solubility of gas in liquidD. Gypsum(iv) HygroscopicE. Silica gel(v) Glucose |
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Answer» A. (v) B. (iii) C. (ii) D. (i) E. (iv) |
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| 348. |
Fill in the blanks: 1. The component present in lesser amount, in a solution is called …….. 2. Example for liquid in solid type solution is ………. 3. Solubility is the amount of solute dissolved in ……… g of solvent. 4. Polar compounds are soluble in ……… solvents. 5. Volume percentage decreases with increases in temperature because ……… |
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Answer» 1. solute 2. amalgam 3. 100 4. Polar 5. of expansion of liquid |
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| 349. |
Choose the odd one out.(a) Air (b) Camphor in nitrogen gas (c) Humid oxygen (d) Salt water |
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Answer» (d) Salt water. a, b and c are gaseous solution whereas d is a liquid solution. |
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| 350. |
Match the following:Column-IColumn-IIA. Blue vitriol(i) CaSO4.2H2OB. Gypsum(ii) CaOC. Deliquescence(iii) CuSO4. 5H2OD. Hygroscopic(iv) NaOH |
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Answer» A. (iii) B. (i) C. (iv) D. (ii) |
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