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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.
| 401. |
In a mixture of 1 gm `H_(2)` and 8 gm `O_(2)`, the mole fraction of hydrogen isA. `0.667`B. `0.5`C. `0.33`D. None of these |
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Answer» Correct Answer - A Mole fraction `=(n)/(n+N)=((w)/(m))/((w)/(m)+(W)/(M))=((1)/(2))/((1)/(2)+(8)/(32))=0.667`. |
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| 402. |
How many gm of `H_(2)SO_(4)` is present in 0.25 gm mole of `H_(2)SO_(4)`A. `24.5`B. `2.45`C. `0.25`D. `0.245` |
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Answer» Correct Answer - A `n=(w)/(m) , w=n xx m =0.25xx98 =24.5 gm` |
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| 403. |
Assertion :- If a liquid solute more volatile than the solvent is added to the solvent, the vapour pressure of the solution may increase i.e., `p_(s)gt p^(@)`. Reason :- In the presence of a more volatile liquid solute, only the solute will form vapours and solvent will not.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C Both the solute and solvent will form the vapour but vapour phase will become richer in the more volatile component. |
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| 404. |
When 6 gm urea dissolve in `18 gm H_(2)O`. The mole fraction of urea isA. `(10)/(10.1)`B. `(10.1)/(10)`C. `(10.1)/(0.1)`D. `(0.1)/(10.1)` |
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Answer» Correct Answer - D Mole fraction `X=(n)/(n + N)=((6)/(60))/((6)/(60)+(180)/(18))=(0.1)/(10.1)`. |
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| 405. |
20 g of a non-volatile solute is added to 500 g of solvent. Freezing point of solvent `= 5.48^(@)C` and solution `= 4.47^(@)C. K_(f)=1.93^(@)//m`. Molecular mas of the solute isA. `72.2`B. `76.4`C. `73.2`D. `70.6` |
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Answer» Correct Answer - B `Delta T_(f)=5.48-4.47=1.01` `Delta T_(f)=K_(f)xx m` `rArr 1.01=1.93xx(20)/(M)xx(1000)/(500)rArr M=76.4`. |
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| 406. |
Freezing point of urea solution is `-0.6^(@)C`. How much urea (M.W. = 60 g/mole) is required to dissolve in 3 kg water `(K_(f) = 1.5^(@)Ckg mol^(-1))`A. 3.6 gB. 2.4 gC. 7.2 gD. 6.0 g |
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Answer» Correct Answer - C `Delta T_(f)=K_(f)xx(1000xxW_(A))/(M_(A)xxW_(B))` [where `W_(A)` = weight of solute, `W_(B)` = weight of solvent, `M_(A)` = molecular weight of solute] or, `0-(-0.6)=(1.5xx1000xxW_(A))/(60xx3000)` or, `W_(A)=(60xx3000xx0.6)/(1.5xx1000)=72g` If we consider the mass of water given to 300 gram `0.6^(@)C=(1000g kg^(-1)xx1.5^(@)Ckg mol^(-1)xx " weight of urea")/(60g mol^(-1)xx300 g)` `therefore` Weight of urea = 7.2 g. |
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| 407. |
Calculate the freezing point of a solution that contains 30 g urea in 200 g water. Urea is a non-volatile, nonelectrolytic solid. `K_(f)` for `H_(2)O=1.86^(@)C//m`A. `4.65^(@)C`B. `-4.65^(@)C`C. `-0.744^(@)C`D. `+0.744^(@)C` |
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Answer» Correct Answer - A `Delta T_(f)=K_(f)xxm=1.86xx(30)/(60)xx(1000)/(200)=4.65^(@)C`. |
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| 408. |
Calculate the molal depression constant of a solvent which has freezing point `16.6^(@)C` and latent heat of fusion `180.75 Jg^(-1)`A. `2.68`B. `3.86`C. `4.68`D. `2.86` |
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Answer» Correct Answer - B `K_(f)=(RT_(f)^(2))/(1000xx L_(f)), R=8.314 JK^(-1)mol^(-1)` `T_(f)=273+16.6=289.6K , L_(f)=180.75 Jg^(-1)` `K_(f)=(8.314xx289.6xx289.6)/(1000xx180.75)=3.86` |
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| 409. |
A solution which has lower osmotic pressure compared to that of other solution is called …….. .A. HypotonicB. HypertonicC. IsotonicD. none of the above |
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Answer» Correct Answer - A A solution with lower osmotic pressure than the other is called hypotonic. |
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| 410. |
If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia complex (which behaves as a strong electrolytes) is `-0.0558^(@)C`, the number of chloride(s) in the coordination sphere of the complex is [`K_(f)` of water `=1.86 Kkgmol^(-1)`] |
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Answer» Correct Answer - 1 `Delta T_(f)=K_(f)xx ixx m rArr 0.0558=1.86xx ixx0.01` i=3 Given complex behaves as a strong electrolyte `alpha = 100%` n = 3(no. of particles given by complex) `therefore` complex is `[Co(NH_(3))_(5)Cl]Cl_(2)` no. of `Cl^(-)` ions in the co-ordination sphere of the complex = 1 `Delta T_(f)=K_(f)xx ixx m` `0.0558=1.86xx ixx0.01` i = 3 |
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| 411. |
The freezing point of water is depressed by `0.37^(@)C` in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed byA. `0.37^(@)C`B. `0.74^(@)C`C. `0.185^(@)C`D. `0^(@)C` |
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Answer» Correct Answer - A The depression in freezing point is proportional to molal concentration of the solute i.e. `Delta T_(f) prop m` `Delta T_(f)=K_(f)mi` or `K_(f)=(Delta T_(f))/(ixxm)` = constant so, `(Delta T_(f_(NaCl)))/(i_(NaCl)xxm_(NaCl))=(Delta T_(f_("Urea")))/(m_("Urea")xxi_("Urea"))` = constant `(0.37)/(2xx0.01)=(Delta T_(f_("Urea")))/(0.02xx1)rArr Delta T_(f_("Urea"))=(0.37xx0.02)/(0.02)=0.37^(@)C`. |
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| 412. |
The ratio of freezing point depression values of `0.01 M` solutions of urea, common salt, and `Na_(2)SO_(4)` areA. `1 : 1 : 1`B. `1 : 2 : 1`C. `1 : 2 : 3`D. `2 : 2 : 3` |
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Answer» Correct Answer - C Urea = 1 , Common salt = 1, `Na_(2)SO_(4)= 3` Ratio = 1 : 2 : 3 |
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| 413. |
Among the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?A. KClB. `C_(6)H_(12)O_(6)`C. `Al_(2)(SO_(4))_(3)`D. `K_(2)SO_(4)` |
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Answer» Correct Answer - c As we know, `DeltaT_(f) = ixxK_(f)xxm` where, `K_(f)` and m are constants. `(a) KCl(aq) overset(DeltaT_(f)propi)(hArr)K^(+)(aq)+Cl^(-)(aq),(i=2)` `(b) C_(6)H_(12)O_(6) hArr No ions` `(c) Al_(2)(SO_(4))_(3)(aq) harr 2 Al^(3+) + 3SO_(4)^(2-), (i=5)` `(d) K_(2)SO_(4)(aq) hArr 2k^(+) + SO_(4)^(2-), (i = 3)` Hence, `Al_(2)(SO_(4))_(3)` will exhibit largest freezing point depression due to the highest value of i. |
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| 414. |
Which of the following solution in water ppossesses the lowest vapour pressureA. 0.1 (M) NaClB. `0.1 (N) BaCl_(2)`C. 0.1 (M) KClD. None of these |
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Answer» Correct Answer - B `BaCl_(2)` gives maximum ion hence it shows lowest vapour pressure. |
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| 415. |
Which of the following solutions in water will have the lowest vapour pressureA. 0.1 M, NaClB. 0.1 M, SucroseC. `0.1 M, BaCl_(2)`D. `0.1 M Na_(3)PO_(4)` |
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Answer» Correct Answer - D `Na_(3)PO_(4)` consist of maximum ions hence it show lowest vapour pressure. `Na_(3)PO_(4)to 3Na^(+)+PO_(4)^(3-)=4` ion. |
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| 416. |
At `25^(@)C`, the total pressure of an ideal solution obtained by mixing 3 mole of A and 2 mole of B, is 184 torr. What is the vapour pressure (in torr) of pure B at the same temperature (Vapour pressure of pure A at `25^(@)C` is 200 torr) ?A. 180B. 160C. 16D. 100 |
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Answer» Correct Answer - B `P=P_(A)+P_(B)=x_(A)p_(A)^(@)+ x_(B)p_(B)^(@)` where P = total pressure, p = partial pressure, `p^(@)` = vapour pressure, x = mole fraction or `184=(3)/(5)xx200+(2)/(5)xx p_(B)^(@)` or, `184 =120+(2)/(5)p_(B)^(@)` or `p_(B)^(@)=160` torr. |
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| 417. |
Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectivelyA. 200 and 300B. 300 and 400C. 400 and 600D. 500 and 600 |
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Answer» Correct Answer - C `p_(A)^(@)+p_(B)^(@)X_(B)=P_("total")` `=p_(A)^(@)xx(1)/(4)+p_(B)^(@)xx(3)/(4)=550` `=p_(A)^(@)+3p_(B)^(@)=550xx4` ………(i) Similarly `=p_(A)^(@)xx(1)/(5)+p_(B)^(@)xx(4)/(5)=560` `p_(A)^(@)+4p_(B)^(@)=560xx5` ……….(ii) Eq. (ii) - Eq. (i) `p_(B)^(@)=560xx5-550xx4=600` So, `p_(A)^(@)=400` `m = 180` |
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| 418. |
Vapour pressure of pure A is 70 mm of Hg at `25^(@)C`. If it forms an ideal solution with B in which mole fraction of A is 0.8 and vapour pressure of the solution is 84 mm of Hg at `25^(@)C`, then the vapour pressure of pure B at `25^(@)C` isA. 140 mmB. 70 mmC. 56 mmD. 28 mm |
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Answer» Correct Answer - a `p = p_(A)^(@) X_(A) + p_(B)^(@) X_(B)` `hArr 84 = 70xx0.8 + p_(B)^(@) xx 0.2` `p_(B)^(@) = (28)/(0.2) = 140 mm` |
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| 419. |
At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mm Hg, respectively. If in a mixture of a and B, the vapoure is 300 mm Hg, the mole fractions of a in the vapour phase, respectively, are -A. `0.6`B. `0.5`C. `0.8`D. `0.4` |
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Answer» Correct Answer - D In equimolar liquid mixture `x_(A)=0.5, X_(B)=0.5` So, `P=0.5xx150+0.5xx100=125` Now let `y_(B)` be the mole fraction of vapour B then `y_(B)=(x_(B)p_(B)^(@))/(P)=(0.5xx100)/(125)=0.4`. |
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| 420. |
If at certain temperature the vapour pressure of pure water is 25 mm Hg and that of a very dilute aqueous urea solution is 24.5 mm Hg, the molarity of the solution isA. `0.02`B. `1.2`C. `1.11`D. `0.08` |
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Answer» Correct Answer - C `(P^(@)-P_(s))/(P^(@))=(w_(B))/(m_(B))xx(m_(A))/(w_(A)), (w_(B))/(w_(B)xx w_(A))xx1000xx(m)/(1000)` `rArr (P^(@)-P_(s))/(P^(@))=" molality " xx(m)/(1000)` `(25-24.5)/(25)=m xx(18)/(1000)` `m=(0.02xx1000)/(18)=1.11` |
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| 421. |
Which statement is true for solution of `0.20 M H_(2)SO_(4)`A. 2 litre of the solution contains 0.020 mole of `SO_(4)^(2-)`B. 2 litre of the solution contains 0.080 mole of `H_(3)O^(+)`C. 1 litre of the solution contains 0.020 mole `H_(3)O^(+)`D. None of these |
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Answer» Correct Answer - B `[H_(3)O^(+)]=2xx0.02=0.04 M` `therefore` 2 litre solution contains 0.08 mole of `H_(3)O^(+)`. |
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| 422. |
`1.0 g` of pure calcium carbonate was found to require `50 mL` of dilute `HCl` for complete reactions. The strength of the `HCl` solution is given by:A. 4 NB. 2 NC. 0.4 ND. 0.2 N |
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Answer» Correct Answer - C M.eq. of HCl = M.eq. of `CaCO_(3)` `N xx50=(1)/(50)xx1000 , N=(1xx1000)/(50xx50)=0.4 N` |
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| 423. |
How much water is to be added to dilute `10 mL` of `10 N HCl` to make it decinormal?A. 990 mlB. 1000 mlC. 1010 mlD. 100 ml |
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Answer» Correct Answer - A `N_(1)V_(1)=N_(2)V_(2)` `10xx10=0.1xx` Volume of new solution Volume of water = 1000 - 10 = 990 ml. |
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| 424. |
The plot of partial vapour vapour pressure of solvent verses its mole fraction in the solution of a constant temp.isA. a straight lineB. a straight line parallel to one axisC. a straight line passing through originD. none of the above |
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Answer» Correct Answer - C `P_("solvent") prop X_("solvent")` |
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| 425. |
The elevation of boiling point method is used for the determination of molecular mass ofA. non-volatile and soluble soluteB. non-volatile and insoluble soluteC. volatile and soluble soluteD. volatile and insoluble solute |
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Answer» Correct Answer - A To show colligative properties solute should be non volatile and soluble in given solvent. |
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| 426. |
Molal elevation constant.A. the elevation in b.pt. which would be produced by dissolving one mole of solute in 100 g of solventB. the elevation of b.pt. which would be produced by dissolving 1 mole solute in 10 g of solventC. elevation in b.pt. which would be produced by dissolving 1 mole of solute in 1000 g of solventD. none |
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Answer» Correct Answer - C `Delta T_(b)=K_(b),`if m=1, m=molality. |
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| 427. |
In which of the following colloidal solution dispersed phase is liquid while dispersion medium is gas?A. dry airB. aerated waterC. amalgamD. moist air |
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Answer» Correct Answer - D Water droplets in air. |
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| 428. |
Which of the following represents a metastable system?A. A dilute solutionB. An unsaturated solutionC. A saturated solutionD. A supersaturated solution |
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Answer» Correct Answer - D A supersaturated solution is metastable because on slight disturbance, solid separates out. |
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| 429. |
Volume of 0.2 `M H_(2)SO_(4)` solution contain 20 millimoles of the solute isA. `10 cm^(3)`B. `100 cm^(3)`C. `20 cm^(3)`D. `200 cm^(3)` |
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Answer» Correct Answer - B 0.2 mol, i.e., 200 millimoles are present in 1000 cc. Hence, 20 milimoles will be present in 100c. |
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| 430. |
The amount of solute (molar mass 60 g `mol^(-1)`) that must be added to 18g of water so that the vapour pressure of water is lowered by 10% isA. 30 gB. 60 gC. 120 gD. 12 g |
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Answer» Correct Answer - B Relative lowering of vapour pressure is given by the formula : `(p^(@)-p_(s))/(p^(@))=(omega_(A))/(m_(A))xx(m_(B))/(omega_(B))` Given, relative lowering of vapour pressure `=(p^(@)-p_(s))/(p^(@))=(10)/(100)` `m_(A)=60, m_(B)=18, omega_(B)=180, omega_(A)=x` `therefore (10)/(100)=(x//60)/(180//18)rArr(1)/(10)=(x//60)/(10)rArr x = 60` Thus, 60 g of the solute must be added to 180 g of water so that the vapour pressure of water is lowered by 10%. |
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| 431. |
36g water and 828g ethyl alcohol form an ideal solution. The mole fraction of water in it, isA. `1.0`B. `0.7`C. `0.4`D. `0.1` |
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Answer» Correct Answer - D `N=(W)/(M)=(828)/(46) = 18, n =(w)/(m)=(36)/(18)=2` `x_(H_(2)O)=(n)/(n+N)=(2)/(2+18)=(2)/(20)=0.1` |
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| 432. |
Which compound corresponds vant Hoff factor (i) to be equal to 2 in dilute solution:A. `K_(2) SO_(4)`B. `NaHSO_(4)`C. SugarD. `MgSO_(4)` |
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Answer» Correct Answer - D i=2 for `MgSO_(4)` as it will lead two ions on complete dissociation. `MgSO_(4) hArr Mg^(2+) + SO_(4)^(-) - (2)` `K_(2)SO_(4) hArr 2K^(+) + SO_(4)^(-) - (3)` `NaHSO_(4) hArr Na^(+) + H^(+) + SO_(4)^(-) - (3)` Sugar rarr sugar (Undissociated) |
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| 433. |
The azeotropic mixture of water (`B.P. = 100^(@)C`) and HCl(`B.P. =86^(@)C`)boils at about `120^(@)C`. During fractional distillation of this mixture it is possible to obtain :A. Pure HClB. Pure waterC. Pure water as well as pure HClD. Neither HCl nor `H_(2)O` in their pure states |
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Answer» Correct Answer - D Azetropic mixture is constant boiling mixture, it is not possible to separate the components of azeotropic mixture by boililng. |
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| 434. |
When an ideal binary solution is in equilibrium with its vapour, molar ratio of the two components in the solution and in the vapur phase is :A. sameB. differentC. may or may not be same depending upon volatile nature of the two componentsD. all |
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Answer» Correct Answer - C Vapour phase composition over liquid phases of mixture may have any value. `P_(A)=P_(0) xx X_(A)` (in vapour phase ) `P_(0)xx X_(A)` (liquid phase ) |
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| 435. |
An aqueous solution containing 1g of urea boils at `100.25^(@)C`. The aqueous solution containing 3 g of glucose in the same volume will boil at (Molecular weight of urea and glucose are 60 and 180 respectively)A. `100.75^(@)C`B. `100.5^(@)C`C. `100.25^(@)C`D. `100^(@)C` |
| Answer» Correct Answer - C | |
| 436. |
The boiling point of benzene is 353.23 K. When 1.80 gm of a nonvolatile silute was dissolved in 90 gm of benzene, the boiling point is raised to 354.11 K. The molar mass of the solute is [`K_(b)` for benzene `= 2.53 K mol^(-1)`]A. `5.8 g mol^(-1)`B. `0.58 g mol^(-1)`C. `58 g mol^(-1)`D. `0.88 g mol^(-1)` |
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Answer» Correct Answer - C The elevation `(Delta T_(b))` in the boiling point `= 354.11 K - 353.23 K = 0.88 K` Substituting these values in expression `M_("Solute")=(K_(b)xx1000xxw)/(Delta T_(b)xx W)` Where, w = weight of solute, W = weight of solvent `M_("solute")=(2.53xx1.8xx1000)/(0.88xx90)=58 gm mol^(-1)` Hence, molar mass of the solute `= 58 gm mol^(-1)` |
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| 437. |
`18 g` of glucose `(C_(6)H_(12)O_(6))` is dissolved in `1 kg` of water in a saucepan. At what temperature will the water boil (at 1 atm) ? `K_(b)` for water is `0.52 K kg mol^(-1)`. |
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Answer» Moles of glucose `=18g//180"g "mol^(-1)=0.1` mol Number of kilograms of solvent `=1kg` Thus molality of glucose solution `=0.1` mol `kg^(-1)` For water, change in boiling point. `DeltaT_(b)=K_(b)xxm=0.52"K kg "mol^(-1)xx0.1" mol "kg^(-1)=0.052K` Since water oils at 373.15K at 1.013 bar pressure, therefore, the boiling point of solution will be `373.15+0.052=373.202K` |
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| 438. |
The boiling a point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. `K_(b)` for benzene is 2.53 K kg `mol^(-1)`. |
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Answer» The elevation `(Deltat_(b))` in the boiling point `=354.11K-353.23K=0.88K` Substituting these values in expression (2.33) we get `M_(2)=(2.53" K kg "mol^(-1)xx1.8" g "xx1000" g "kg^(-1))/(0.88Kxx90g)=58" g "mol^(-1)` |
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| 439. |
A solution of glucose in water is labelled as `10 "percent" w//w`, what would be the molality and mole fraction of each component in the solution? If the density of the solution is `1.2 g mL^(-1)`, then what shall be the molarity of the solution? |
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Answer» 10 percent w/w solution of glucose I water means 10 g glucose and 90 g of water. Molar mass of glucose`=180g" "mol^(-1)` and molar mass of water `=18" "mol^(-1)`. `10g` of glucose `=(10)/(180=0.0555` moles and 90g of `H_(2)O=(90)/(18)=5` moles `therefore` molality of solution `=("Moles of solute"xx1000)/("Mass of solution in grams")` `=(0.555)/(90)xx1000=0*617`m Mole fraction of glucose. Mole fraction of glucose. `=X_(g)=("No of moles of glucose")/("No. of moles+No. of moles of glucose of water")` `=(0*0555)/(5+0*0555)=0.01` mole fraction of water `X_(w)=("No. of moles of water")/("No. of moles+No. of glucose moles of water")` `=(5)/(5+0*0555)=0*99` Volume of 100g of solution `=("Mass of solution")/("Density")=(100)/(1*2)=83*33mL` `therefore` Molarity of solution `=(0*0555)/(83*33)xx1000` `=0*67M`. |
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| 440. |
Volume of `0.1M HCl` required to react completely with 1g equimolar mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` is |
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Answer» Calculate of no. of moles of components in the mixture Let x g of `Na_(2)CO_(3)` is present in the mixture. `therefore(1-x)g` of `NaHCO_(3)` is present in the mixture. Molar mass of `Na_(2)CO_(3)` ,brgt `=2xx23+12+3xx16=106" g "mol^(-1)` and molar mass of `NaHCO_(3)` `=23xx1+1+12+3xx16=84" g "mol^(-1)` No. of moles of `Na_(2)CO_(3)` in x g`=(x)/(106)` No. of moles of `NaHCO_(3)` in (1-x)`g=(1-x)//84` As given that the mixture contains equimolar amounts of `Na_(2)CO_(3)` and `NaHCO_(3)` therefore. `(x)/(106)=(1-x)/(84)` `=106-106x=84x` `106=190x` `thereforex(106)/(190)=0*558g` `therefore` No. of moles of `Na_(2)CO_(3)` present `=(0*558)/(106)=0*00526` and no. of moles of `NaHCO_(3)` present `=(1-0*558)/(84)=0*00526` Calculation of no. of moles of of HCl required `Na_(2)CO_(3)+2HClto2NaCl+H_(2)O+CO_(2)` `NaHCO_(3)+HClto NaCl+H_(2)O+CO_(2)` As can be seen, each mole of `Na_(2)CO_(3)` needs ltBrgt 2 moles of HCl, `therefore0*00526` mole of `Na_(2)CO_(3)` needs `=0*00526xx2=0*01052` moles Each mole of `NaHCO_(3)` needs 1 mole of HCl. `0*00526=0*00526` mole Total amount of HCl needed will be `=0*01052+0*00526=0*01578` mole `0*1` mole of 0.1 M HCl are present in 1000 mL of HCl `therefore0*01578` mole of `0*1` M HCl will be present n `=(1000)/(0*1)xx0*01578=157*8mL` |
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| 441. |
A solution of urea contain 8.6 gm/litre (mol. wt. 60.0). If is isotonic with a 5% solution of a non-volatile solute. The molecular weight of the solute will beA. `348.9`B. `34.89`C. 3489D. `861.2` |
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Answer» Correct Answer - A For two non-electrolytic solution if isotonic, `C_(1)=C_(2)` `therefore (8.6)/(60xx1)=(5xx1000)/(m.wt. xx 100) therefore m = 348.9` |
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| 442. |
`K_(f) of 1,4-"dioxane is" 4.9mol^(-1)` for `1000g`.The depression in freezing point for a`0.001m` solution in dioxane isA. `0.0049`B. `4.9+0.001`C. `4.9`D. `0.49` |
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Answer» Correct Answer - A `Delta T=K_(f)xx` Molality `= 4.9xx0.001=0.0049 K` |
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| 443. |
The solution containing 4.0 gm of a polyvinyl chloride polymer in 1 litre of dioxane was found to have an osmotic pressure `6.0xx10^(-4)` atmosphere at 300K, the value of R used is 0.082 litre atmosphere `"mole"^(-1)K^(-1)`. The molecular mass of the polymer was found to beA. `3.0xx10^(2)`B. `1.6xx10^(-5)`C. `5.6xx10^(4)`D. `6.4xx10^(2)` |
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Answer» Correct Answer - B `pi V=(w)/(m)RT` `therefore 6xx10^(-4)xx1=(4)/(m)xx0.0821xx300 , m = 1.64xx10^(5)` |
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| 444. |
osmotic presure of a sugar solution at `27^(@)C` is 2.46 atmosphere. The concentration of the solution in mol per litre is :A. 1 MB. 0.01 MC. 0.0125 MD. 0.1 M |
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Answer» Correct Answer - D `pi V = nRT, pi = C R T ` `:. C= pi/(RT)= 2.46/(0.082 xx (27 + 273))=0.1 M` |
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| 445. |
The solution containing `4.0gm` of a polyvinyl chloride polymer in `1` litre dioxane was found to have an osmotic pressure `6.0xx10^(-4)`atmosphere at `300K`,the value of R used is `0.082`litre atmosphere`"mole"^(-1)K^(-1)`.The molecular mass of the polymer was found to beA. `3 xx 10^(3)`B. `1.6 xx 10^(5)`C. `5 xx 10^(4)`D. `6.4 xx 10^(2)` |
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Answer» Correct Answer - B `piV= W_(2)/M_(2) RT,` `:. 6 xx 10^(-4) xx 1 = 4/M_(2) xx 0.0821 xx 300,` `M_(2) = 1.64 xx 10^(5)` |
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| 446. |
The osomotic pressure of a solution at `0^(@)C` is `4 atm`. What will be its osmotic pressure at `546 K` under similar conditions? a.`4 atm`, b.`9 atm`,c.`8 atm`, d.`6 atm`A. 0.5 atmB. `2 xx 273 `atmC. 4 atmD. `273 // 2` atm |
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Answer» Correct Answer - C `pi_(1)/pi_(2)=T_(1)/T_(2), :. pi_(1)/2=546/273, pi_(1)=4 "atm"` |
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| 447. |
The osomotic pressure of a solution at `0^(@)C` is `4 atm`. What will be its osmotic pressure at `546 K` under similar conditions? a.`4 atm`, b.`9 atm`,c.`8 atm`, d.`6 atm`A. 4 atm.B. 2 atmC. 8 atm.D. 1 atm. |
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Answer» Correct Answer - C `pi prop T.` If T is doubled, P is also doubled. |
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| 448. |
The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1g of urea in the same quantity of water is :A. 1 gB. 3 gC. 6 gD. 18 g |
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Answer» Correct Answer - B `(P^(@)-P_(s))/(P_(s))=(w_(1)M_(2))/(w_(2)M_(1))` To produce same lowering of vapour pressure `((P^(@)-P_(s))/(P_(s)))` term will be same for both the case (solvent is same) `therefore (w_(g)xx18)/(50xx180)=(w_(u)xx18)/(50xx60)` (`w_(g)` = weight of glucose, `w_(u)` = weight of urea) or, `(w_(g)xx18)/(50xx180)=(1xx18)/(50xx60)` or, `w_(g)=3`. |
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| 449. |
The relative lowering of vapour pressure in case of dilute solution is directly proportional to:A. molalityB. molarityC. mole fractionD. all |
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Answer» Correct Answer - D In case of dilute solution `(P_(1)^(0)-P)/P_(1)^(0)` prop all . |
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| 450. |
Which has minimum osmotic pressureA. 200 mL of 2 M NaCl pressureB. 200 mL of 1 M glucose solutionC. 200 mL of 2 M urea solutionD. All have same osmotic pressure |
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Answer» Correct Answer - B Osmotic pressure is a colligative property. Particles in NaCl solution `= 2xx2 = 4M` . Particles in glucose solution `= 1xx1 =1M`. Particles in urea solution `= 2xx1=2 M`. Least particles are in glucose solution, hence its osmotic pressure is the minimum. |
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