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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.
| 601. |
Blood cells retain their normal shape in solution which areA. hypertonic to bloodB. isotonic to bloodC. hypotonic to bloodD. isoelectronic to blood |
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Answer» Correct Answer - 2 In isotonic solution, red blood cells (RBC) neither swell nor undergo plasmolysis. In hypotonic solution (i.e less concentrated solution), RBC swell and even burst. In hypertonic solution (i.e more concentration solution), RBC shrink due to plasmoysis. |
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| 602. |
When a solution of `CHCI_(3)` is mixed with a solution of acetone,`DeltaV_("mix")` isA. PositiveB. NegativeC. ZeroD. Cannot be predicted |
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Answer» Correct Answer - D Mass of solvent does not change with temperature. |
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| 603. |
A liquid mixtue having composition corresponding to point z in the figure shown is subjected to distillation at constant pressure. Which of the following statement is correct about the process. A. The composition of distillate differs from the mixtureB. The boiling point goes on changingC. The mixture has lowest vapour pressure then for any other compositionD. Composition of an azeotrope alters on changing the external pressure. |
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Answer» Correct Answer - C::D Azoeotropic composition depends on external pressure. |
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| 604. |
Blood is isotonic with:A. `0.16 M NaCI`B. Conc. `NaCI`C. `50% NaCI`D. `30% NaCI` |
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Answer» Correct Answer - A Blood is isotomic with `0.16M NaCI` |
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| 605. |
Blood is isotonic with:A. `0.16 M NaCI`B. Conc. `NaCI`C. `30% NaCI`D. `50% NaCI` |
| Answer» Correct Answer - A | |
| 606. |
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 adn water mixtures 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 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 `= 155.7 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 answering 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^(s) = (P^(@))_(solvent). X_(solvent)` `= 40 xx 0.9` `= 36.0 mm Hg` |
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| 607. |
The freezing point (`"in"^(@)"C"`) of a solution containing 0.1 g of `K_(3)[Fe(CN)_(6)]` ( Mol. Wt. 329) in 100g of water (`K_(f)"=1.86 k kg mol"^(-1))` isA. `-2.3xx10^(-2)`B. `-5.7xx10^(-2)`C. `-5.7xx10^(-3)`D. `4.0xx10^(-6)` |
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Answer» Correct Answer - A `"For" K_(3)[Fe(CN)_(6)]i=4` `DeltaT_(f)=ixx(K_(f)W_("solute")xx1000)/(m_("solute")xxw_("solvent"))` `DeltaT_(f)=4xx1.86xx(0.1xx1000)/(329xx100)` `thereforeT_(f)("solution")=0.0023=-2.3xx10^([email protected])"C"` |
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| 608. |
The freezing point (in `.^(@)C)` of a solution containing `0.1g` of `K_(3)[Fe(CN)_(6)]` (Mol. wt. `329`) in `100 g` of water `(K_(f) = 1.86 K kg mol^(-1))` is :A. `-2.3 xx 10^(-2)`B. `-5.7 xx 10^(-2)`C. `-5.7 xx 10^(-3)`D. `-1.2 xx 10^(-2)` |
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Answer» Correct Answer - A `K_(3)[Fe(CN)_(6)]rarr3K^(+)+[Fe(CN)_(6)]^(3-)` `i=4` `DeltaT=ixxK_(f)xx(w_(B)xx1000)/(m_(B)xxw_(A))=4xx1.86xx(0.1xx1000)/(329xx100)` `=2.3xx10^(-2)` `:.` Freezing point of solution will be `-2.3 xx 10^(-2) .^(@)C`] |
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| 609. |
The freezing point (in `.^(@)C)` of a solution containing `0.1 g` of `K_(3)[Fe(CN)_(6)]` (Mol.wt. 329) in 100 g of water `(K_(f) = 1.86 K kg mol^(-1))` isA. `-2.3 xx 10^(-2)`B. `-5.7 xx 10^(-2)`C. `-5.7 xx 10^(-3)`D. `-1.2 xx 10^(-2)` |
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Answer» Correct Answer - A `DeltaT_(f) = m xx K_(f) xx i` `=(0.1//329)/(100) xx 1000 xx 1.86 xx 4 = 0.2261` `= 2.3 xx 10^(-2)` `T_(f) = 0 - DeltaT_(f) =- 2.3 xx 10^(2).^(@)C` |
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| 610. |
`0.5`normal sugar solution is isotonic withA. `1.0`normal glucose solutionB. `1.2`normal potassium chloride solutionC. `0.5`normal urea solutionD. `5% urea solution |
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Answer» Correct Answer - C For isotonic solution,`C_(1)=C_(2)` |
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| 611. |
Mixture of two liquids A and B is placed in cylinder containing piston. Piston is pulled out isotehrmally so that volume of liquid decreases but that of vapour increases. When negligibly small amount of liquid was remaining the mole fraction of A in vapour is `0.4`. Given `P_(A)^(@) = 0.4` atm and `P_(B)^(@) = 1.2` atm at the experimental temperature. Calculate the total pressure at whcih the liquid has almost evaporated. (Assume ideal behaviour) |
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Answer» Correct Answer - `0.66 atm` `P_(tau)=P_(A)^(@)X_(A)P_(B)^(@)X_(B) = P_(A)^(@) X_(A) +P_(B)^(@) (1-X_(A))` `P_(tau) = P_(B)^(@) +X_(A) (P_(A)^(@) -P_(B)^(@))` `Y_(A) = (P_(A)^(@) xx A)/(P_tau) = (P_(A)^(@)-A)/(P_(B)^(@)+X_(A)(P_(a)^(@) -P_(B)^(@)))` `0.4 = (0.4X_(A))/(1.2 =0.8 X_(A))` `1.2 = 1.8 xx A` `X_(A) = (2)/(3)` so `(1)/(3)` `P_(tau) = 0.4 xx (2)/(3) +1.2 xx (1)/(3) = (2)/(3) = 0.66` atm |
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| 612. |
Two solutions have different osmotic pressures. The solution of higher osmotic pressure is called:A. isotomic solutionB. hypotonic solutionC. isotopic solutionD. hypertonic solution |
| Answer» Correct Answer - D | |
| 613. |
If `20 mL` of ethanol (density `=0.7893g//mL)` is mixed with `40mL` water (density `= 0.9971g//mL)` at `25^(@)C`, the final slution has density of `0.9571g//mL`. Calculate the percentage change in total volume of mixing. Also calculate the molarity of alcohol in the final solution. |
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Answer» Correct Answer - `3.1 % 8.6` `C_(2)H_(5)OHrarrV_(1)=20 mL, d_(1)=0.7893 g//mL` `m_(1)=15.786 g` `H_(2)Orarr V_(2)=40 mL, d_(2) = 0.9971 g//mL`. `m_(2)=39.884 g` Total mass `= 55.65 g` `d_(sol)=0.9571 g//mL` `V_(sol) = 58.14 mL` % change `= (60-58.14)/(60)xx100 =3.1%` `m = (15.766xx1000)/(46xx39.884)=8.6` |
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| 614. |
Solutions having the same osmotic pressure are called:A. isotonic solutionB. molar solutionsC. hypotonic solutionsD. ideal solutions |
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Answer» Correct Answer - A By definition |
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| 615. |
The addition of `3g` of a substance to `100g C CI_(4)(M = 154 g mol^(-1))` raises the boiling point of `C CI_(4)` by `0.60^(@)C` if `K_(b)(C CI_(4))` is `5K mol^(-1) kg`, calculate (a) the freezing point depression (b) the relative lowering of vapour pressure (c) the osmotic pressure at `298K` and (d) the molar mass of the substance. Given: `K_(f) (C CI_(4)) = 31.8 K kg mol^(-1)` and `rho` (solution) `= 1.64 g cm^(-3)` |
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Answer» Correct Answer - (a) `3.816k`, (b) `01814`, (c) `4.669 atm` (d) `250 g mol^(1-)` (a) `(DeltaT_(f))/(DeltaT_(b)) = (K_(f))/(K_(b))` `DeltaT_(f) = (31.8)/(5.03) xx 0.6 = 3.79^(@)C` (b) `m = (DeltaT_(b))/(K_(b)) = (0.6)/(5.03) = (x xx1000)/((1-x)154)` `rArr 1-x = 54.44 x` `rArr 55.44 x = 1` `x = (DeltaP)/(P^(@)) = (1)/(5.44) = 0.018` (c) `(1)/(55.44) = ((3)/(M_(1)))/((3)/(M_(1))+(100)/(154))` `rArr (100)/(154) = (166.32)/(M_(1)) M_(1) = 231.5` `M = (0.6)/(5.03)` `(M xx 1000)/(1000 xx 1.64 -M xx 251.5)` `1640 = 8634.83M` `M = 0.19` `p = n//V RT = 0.19 xx 0.0821 xx 298` `=4.65 atm` |
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| 616. |
The addition of `3g` of a substance to `100g C CI_(4)(M = 154 g mol^(-1))` raise the boiling point of `C CI_(4)` by `0.60^(@)C` of `K_(b)(C CI_(4))` is `5.03 kg mol^(-1) K`. Calculate: (a) the freezing point depression (b) the relative lowering of vapour pressure (c) the osmotic pressure at `298K` the molar mass of the substance Given `K_(f)(C CI_(4)) = 31.8 kg mol^(-1) K` and `rho` (density) of solution `= 1.64g//cm^(3)` |
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Answer» Correct Answer - (a) `3.793^(@)C` (b) `0.018`, (c) `4.65 atm`, (d) `251.5` (a) `(Delta T_(t))/(Delta T_(b)) = (k_(f))/(k_(b))=Delta T_(f)=(0.6xx31.8)/(5.03)=3.793^(@)C` (b) Relative lowering of vapour pressure = `(n)/(n+N)=((3)/(251.5))/((3)/(251.5)+(100)/(154))=0.018` (c ) `pi = CRT` `n = (3)/(251.5)=0.012` `v = (103)/(1.64) = 62.8 mL` `pi = (0.012)/(0.0628)xx0.0821xx298=4.65` atm (d) `0.6 = (5.03xx3xx1000)/(M_(W)xx100) = gt M_(W) = 251.5` |
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| 617. |
When `0.6g` of urea dissolved in `100g` of water, the water will boil at `(K_(b)` for water `= 0.52kJ. mol^(-1)` and normal boiling point of water `=100^(@)C)`:A. `373.052 K`B. `273.52K`C. `372.48K`D. `273.052K` |
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Answer» Correct Answer - A `DeltaT_(b) = ((0.6)/(60))/(100) xx 1000 xx 0.52` `= (1)/(1000) xx 1000 xx 0.52 = 0.52` `T - 373 = 0.52 T = 373.52` |
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| 618. |
Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K. |
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Answer» P0Pure benzene = 50.71 mm Hg P0nepthalene = 32.06 mm Hg Number of moles of benzene = \(\frac{39}{78}\) = 0.5 mol Number of moles of naphthalcne = \(\frac{128}{128}\) = 1 mol Mole fraction of benzene = \(\frac{0.5}{1.5}\) = 0.33 Mole fraction of naphthalene = 1 – 0.33 = 0.67 Partial vapour pressure of benzene =P0benzene x (Mole fraction of benzene) = 50.71 x 0.33 = 16.73 mm Hg Partial vapour pressure of naphthalene = 32.06 x 0.67 = 21.48mm Hg Mole fraction of benzene in vapour phase = \(\frac{16.73}{16.73+21.48}\) = \(\frac{16.73}{38}\) = 0.44 Mole fraction of naphthalene in vapour phase = 1 – 0.44 = 0.56 |
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| 619. |
The relative lowering in vapour pressure is proportional to the ratio of number ofA. solute molecules to solvent moleculesB. solvent molecules of solute moleculesC. solute molecules to the total number of molecules in solutionD. solvent molecules to the total number of molecules in solution |
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Answer» Correct Answer - C `(p^@-p_s)/p^@=n_2/(n_1+n_2)` |
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| 620. |
If the vapour pressure of pure A and pure B at 298 K are 60 and 15 torr respectively, what would be the mole fraction of A in vapour phase (at this temperature) in a solution that contains 20 mole per cent of A in the (A + B) binary mixture in liquid phase ?A. 0.2B. 0.3C. 0.5D. 0.7 |
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Answer» Correct Answer - C `x_(A)=(20)/(100)=0.2,:.x_(B)=0.8` `p=p_(A)+p_(B)=p_(A)^(0)x_(A)+p_(B)^(0)x_(B)` `= 60xx0.2+15xx0.8=12+12=24` Mole fraction of A in the vapour phase, i.e., `y_(A)` may be calculatd as : `p_(A)=y_(A)xxp` `12=y_(A)xx24` `y_(A)=(12)/(24)=0.5`] |
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| 621. |
An aqueous solution of acetone, `CH_(3)COCH_(3)` is `10.00%` acetone by weight. What is the mole percentage of acetone in this solution:A. `3.332%`B. `5.000%`C. `10.00%`D. `11.11%` |
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Answer» Correct Answer - A Mole fraction `= ((10)/(58))/((10)/(50)+(90)/(18)) = 0.0332` `%` mole `=3.32%` |
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| 622. |
The freezing poing of an aqueous solution of a non-electrolyte is `-0.14^(@)C`. The molarity of this solution is `[K_(f) (H_(2)O) = 1.86 kg mol^(-1)]`:A. `1.86m`B. `1.00m`C. `0.15m`D. `0.075m` |
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Answer» Correct Answer - D `0.14 = 1.86 xx m` `m = 0.075` |
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| 623. |
0.1 formal soluton of NaCl is found to be isotonic with 1.10% solution of urea. Calculate the apparent degree of ionization of NaCl. |
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Answer» Correct Answer - 0.83 `pi_(NaCl)=pi_(urea)` `=nxx(1)/(V)RT` `ixx0.1xxRT=(1.1)/(60xx0.1)xxRT" "(pi=(n)/(V)RT)` `i=83` `alpha=(i-1)/(n-1)=(1.83-1)/(2-1)=0.83`] |
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| 624. |
`(0.1M "urea")/((A)),(0.1M NaCl)/((B)),(0.1MBaCl_(2))/((C))` |
| Answer» `{:("order of" pi,CgtBgtA),("order of" R.L.V.P,C gt B gt A),("order of" V.P,Agt B gt C),("order of" DeltaT_(B),C gt B gt A),("order of" T_(B) "of solution", C gt B gt A),("order of" DeltaT_(F),C gt B gt A),("order of" T_(F) " of solution",A gt B gt C):}` | |
| 625. |
An ionic compound that attracts atmospheric water so strongly that a hydrate is formed is said to be :A. DiluteB. HygroscopicC. ImmiscibleD. Miscible |
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Answer» Correct Answer - B Certain compounds combine with the moisture os atmosphere and are converted into hydroxides or hydrates. Such substances are called hygroscopic. `e.g.`, anhydrous `CuSO_(4)`, quick lime `(CaO)`, anhydrous `Na_(2)CO_(3)` etc. |
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| 626. |
Which of the following relations is(are) correct according to Freundlich? (P)x/m = constant , (Q) x/m = constant `xxp^(1//n)(ngt1),` (S) x/m = constant `xxp^(n)(ngt1)`A. All are correctB. All are wrongC. (ii) is correctD. (iii) is correct |
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Answer» Correct Answer - C According to Freundlich adsorption isotherm, `x/m prop kp^(1//n) (n gt 1)`. |
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| 627. |
Explain abnormal molecular masses. |
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Answer» When the observed molecular masses obtained from their colligative properties of the substances are different (higher or lower) than the theoretical or normal values calculated from their molecular formulae, then they are called abnormal molecular masses. |
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| 628. |
Abnormal molar masses arises due toA. association or dissociation of moleculesB. change in number of moleculesC. Both a & bD. Due to increase in concentration of solution |
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Answer» Correct Answer - C |
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| 629. |
Abnormal molar mass is produced by : (a) association of solute (b) dissociation of solute (c) both association and dissociation of solute (d) separation by semipermeable membrane |
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Answer» Option : (c) both association and dissociation of solute |
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| 630. |
Solubility of gas decreases in a liquid byA. Increase of temperatureB. CoolingC. Increasing pHD. Decreasing pH |
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Answer» Correct Answer - 1 |
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| 631. |
The temperature at which the vapour pressure of a liquid becomes equals to the external (atmospheric) pressure is itsA. Freezing pointB. Boiling pointC. Melting pointD. Critical temperature |
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Answer» Correct Answer - 2 |
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| 632. |
The temperature at which the vapour pressure of a liquid becomes equals to the external (atmospheric) pressure is itsA. melting pointB. sublimation pointC. inversion pointD. boiling point |
| Answer» It is known as b.pt | |
| 633. |
What are the types of Non-Ideal solutions? |
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Answer» Types of Non-Ideal solutions: (i) Non-ideal solution showing positive deviations (ii) Non-ideal solution showing negative deviations |
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| 634. |
A `100 cm^(3)` solution is prepared by dissolving 2g of `NaOH` in water. Calculate the normality of the solution. |
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Answer» `2g NaOH = (2)/(40) g .eq = (1)/(20)g eq.` `N = ((1)/(20))/(100) xx 1000 = (1)/(2) = 0.5N` |
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| 635. |
Why does the use of pressure cooker reduce the cooking time? |
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Answer» At higher pressure over the liquid, the liquid boils at high temperature. Therefore, cooking occur fast. |
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| 636. |
Two liquids A and B are mixed and the resulting solution is found to be cooler. What do you conclude about the deviation from ideal behaviour? |
| Answer» The solution shows positive deviation from ideal behaviour. | |
| 637. |
Why is the vapour pressure of liquid constant at a constant temperature ? |
| Answer» Vapour pressure is the pressure of the vapour at equilibrium state when the rate of evaporation becomes equal to the rate of condensation. The equilibrium constant does not change at a particular temperature and therefore the vapour pressure remains constant. | |
| 638. |
Two liquid A & B are mixed and the resulting solution is found to be cooler. What do you conclude about from ideal behavior? |
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Answer» Positive deviation. |
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| 639. |
If to a solvent (vapour pressure `=p^(@))`, n moles of a non volatile solute are dissolved then the vapour pressure of the solution p isA. `p^(@) -p = p^(n)`B. `p =p^(@)(1-n)`C. `p= p^(@)n`D. None of the above |
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Answer» Correct Answer - D Partial vapour pressure of a component in a solution cannot be obtained without knowing the number of moles of both solvent and solute present in solution. |
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| 640. |
The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm3) |
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Answer» Let the solution be 1 litre or 1000 cm3 ∴ Number of moles of H2SO4 = 1.35 Wt. of solution = 1000 x 1.02 = 1020 g Wt. of sulphuric acid = 1.35 x 98 = 132.3g Wt. of water = 1020 - 132.3 = 887.79 Molality of H2SO4 = ppp x 1000 = 1.52 m From (i) M1 = 110.82, from (ii) M2 = 196.15 AB4 – AB, = B2 196.15 - 110.82 = B2 85.33 = B2 B = 42.665 Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B 110.82 = Atomic mass of A + 85.33 Atomic mass of A = 110.82 - 85.33 = 25.49 Atomic mass of A = 25.499 Atomic mass of B = 42.669 |
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| 641. |
Converting mass percent to molarity: the density of a 24.5 mass % solution of sulphuric acid `(H_(2)SO_(4))` in water is `1.176 g mL^(-1)` at `25.0^(@)C`. What is the molarity of the solution? Stregy: Use mass % to calculate number of moles of solute and use density of solution to calculate volume of solution. Describing a solution as 24.5 mass % `H_(2)SO_(4)` in `H_(2)O` means that every 100.0 g of solution contains 24.5 g of `H_(2)SO_(4)` and 75.5 g of `H_(2)O`. Since we want to calculate the concentration in molarity, we first need to find the number of moles of `H_(2)SO_(4)` dissolved in a specific mass of solution. Next we use density as a conversion factor to find the volume of that solution and then calculate molarity by dividing the number of moles of `H_(2)SO_(4)` by the volume of solution. |
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Answer» Consider `100.0 g` of solution. Therefore, mass of `H_(2)SO_(4)` is `24.5 g`. First, convert the mass of `H_(2)SO_(4)` into moles. `n_(H_(2)SO_(4))=(Mass H_(2)SO_(4))/(Molar mass H_(2)SO_(4))` `=(24.5 g H_(2)SO_(4))/(98.0 g H_(2)SO_(4)//mol H_(2)SO_(4))` `=0.25 mol H_(2)SO_(4)` Next, we find the volume of `100.0 g` of solution using sensity as the conversion factor: Volume of solution`=((1216g)/(1000mL))/((100.0" g soln")/(1.176"g soln"//mL))` `=85mL` `=85 mL` Finally, calculate the molarity of solution using Eqution `2.14`. Molarity`=n_(H_(2)SO_(4))/V_(mL)xx(1000 mL)/L` `=(0.25 mol H_(2)SO_(4))/(85 mL soln)xx(1000 mL)/L` `=2.94 M` Thus, the molarity of the `24.5` mass `%` of sulphuric acid solution is `2.94 M`. Short cut method: We can also convert mass percent into molarity by using the following formula: `M=("(mass%)(specific gravity or density of soln)(10)")/(("molar mass of solute"))` `=((24.5)(1.175)(10))/98` `=2.94 mol L^(-1)` |
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| 642. |
The following graph represents variation of boiling point with composition of liquid and vapours of binary liquid mixture. The graph is plotted at constant pressure. Which of the following statement(s) is incorrect. Here X & Y stands for mole fraction in liquid and vapour phase respectively. A. `X_("benzene") = 0l.5` and `Y_("toluene") = 0.2`B. `X_("toluene") = 0.3` and `Y_("benzene") = 0.6`C. `X_("benzene") = 0.3` and `Y_("toluene") = 0.4`D. `X_("benzene") = 0.7` and `Y_("toluene") = 0.3` |
| Answer» Correct Answer - A::C | |
| 643. |
What is the vapour pressure of a solution of glucose which has an osmotic pressure of 3 atmosphere at `20^(@)C` ? The vapour pressure of water at `20^(@)C` is 17.39 mm. Consider the density of solution equal to that of solvent. |
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Answer» Correct Answer - 17.35 mm Use : `(Deltap)/(p_(0))=(pim)/(dRT)`]. |
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| 644. |
The solution which has higher osmotic pressure than some other solution is known as……… |
| Answer» Correct Answer - C | |
| 645. |
At a freezing point, (a) Vapour pressure of a solution = Vapour pressure of a solid (b) Vapour pressure of a liquid = Vapour pressure of a solid (c) Vapour pressure of a liquid > Vapour pressure of a solid (d) Vapour pressure of a solid > Vapour pressure of a liquid |
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Answer» Option : (b) Vapour pressure of a liquid = Vapour pressure of a solid. |
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| 646. |
As temperature increases :(a) Osmotic pressure and vapour pressure decrease (b) Vapour pressure and osmotic pressure increase (c) Vapour pressure increases but osmotic pressure decreases (d) Osmotic pressure increases but vapour pressure decreases |
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Answer» Option : (b) Vapour pressure and osmotic pressure increase |
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| 647. |
Vapour pressure of a liquid depends upon itsA. temperatureB. surface areaC. nature of liquidD. external pressure |
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Answer» Correct Answer - A::C Vapour pressure of a liquid depends only upon temperature and nature of the liquid. |
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| 648. |
In osmosis :(a) solvent molecules pass from high concentration of solute to low concentration(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute(c) solute molecules pass from low concentration to high concentration(d) solute molecules pass from high concentration to low concentration |
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Answer» Option : (b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute. |
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| 649. |
In istonic solutions………..A. solute and solvent both are sameB. osmotic pressure is sameC. solute and solvent may or may not be sameD. solute is always same solvent may be different |
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Answer» Correct Answer - B::C The two solution having same osmotic pressure are known as isotonic solution. The solute and solvent particles may or may not same but osmotic pressure must be same. |
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| 650. |
Mole fraction of a solution in 1.00 molal aqueous solution isA. `0.1770`B. `0.0177`C. `0.0344`D. `0.83` |
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Answer» Correct Answer - B For electrolytes, `Delta T_(f)=ixxk_(f)xxm` (min molarity) or `Delta T_(f)=ixxk_(f)xx(W_(2))/(M_(2))xx(1000)/(W_(1)), Delta T_(f)=3.82^(@)C` `k_(f)=1.86 K kg mol^(-1), M_(2)=142` for `Na_(2)SO_(4)` `W_(2)=5g, W_(1)=45.0g` `therefore 3.82=ixx1.86xx(5)/(142)xx(1000)/(45)` `therefore i=(3.82xx142xx45)/(1.86xx5xx1000)=2.63` |
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