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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.
| 651. |
If two liquids A and B form minimum boiling azetrope at some spectific composition, thenA. A-B ineractions are stronger than those between A-A of B-B interactions.B. Vapour pressure of solution increases because more mumber of molecules of liquids A and B can escape from the solution.C. vapour pressure of solution decreses because less number of molecules of only one of the liquids escape from the solution.D. A-B interactions are weaker than those between A-A or B-B. |
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Answer» Correct Answer - d if two liquids on mixing are to form a minimum boiling azeotrope, the vapour pressure of the solution formed must be maimum (boiling point will be minimum). This can happen if A -B interactions are weaker than both A-A and B-B interactions. |
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| 652. |
The system that forms maximum boiling points azetrope isA. carbon disulphide acetoneB. benzene tolueneC. acetone-chloroformD. n-hexane-n-heptane |
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Answer» Correct Answer - C Mole fraction of solute `=("Moles of solute")/("Moles of solute + Moles of solvent")` `= (1)/(1+(1000)/(18))=(1)/(1+55.56)=0.0177` |
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| 653. |
Assertion : In an ideal solution, `DeltaH` maximum is zero. Reson A…B interaciton are the same as A…A and B…B interations.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 - a Reason is the correct explanation for assertion. |
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| 654. |
The attractive forces between molecules of a molecular crystal areA. relatively weak dipole-dipole intractionsB. relatively weak dispersion forcesC. strong hydrogen bondingD. any of these |
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Answer» Correct Answer - 4 Consider naphthalene `(C_(10)H_(8))` as an example. In a naphthalene crystal the nonpolar `C_(10)H_(8)` molecules are held together solely by dispersion forces. Thus we can predict that naphthalene should dissolve readily in nonpolar solvents, such as benzene, but only slightly in water. This is indeed the case. On the other hand, we find that urea `(H_(2)NCONH_(2))` is much more soluble in water and ethanol than in `C Cl_(4)` or `C_(6)H_(6)`, because it has dipole-dipole interactions, intermolecular hydrogen bonding and the capacity to form hydrogen bonds with water. (The `H` atoms and the `O` and `N` atoms in urea can form hydrogen bonds with `H_(2)O` and `C_(2)H_(5)OH` molecules) Similarly, if we compare the solubilities of molecular `I_(2)`, a nonpolar substance, in water and in carbon tetrachloride, we notice that `I_(2)` is much more soluble in `C Cl_(4)` than in `H_(2)O`. |
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| 655. |
Consider the two figures given below : Which of the following statements regarding the experiment is true ?A. The solubility of a gas in liquid in beaker (i) is greater than that is beaker (ii)B. The solubility of a gas in beaker (i) is less than that is beaker (ii)C. The solubility of a gas is equal in both beakers .D. The solubility of a gas remains unaffected by change in weights . |
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Answer» Correct Answer - B The solubility of gas in a liquid increases with increase in pressure and is directly proportional to the pressure of the gas. |
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| 656. |
At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads toA. low concentrations of oxygen in the blood and tissuesB. high concentrations of oxygen in the blood and tissuesC. release of dissolved gases and formation of bubbles of nitrogen in the bloodD. thickening of blood and tissues |
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Answer» Correct Answer - A At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. |
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| 657. |
Distribution coefficient of an organic acid between water and benzene is 4.1 in favour of `C_(6)H_(6)`. If 5g of acid is distributed in between 50mL of benzeen and `100mL` of water, calculate the concentration of acid in two solvents. |
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Answer» Let the amount of organic acid in `C_(6)H_(6)` layer `= ag` volume of `C_(6)H_(6) = 50 mL` `:.` concentration of acid in `C_(6)H_(6) = (a)/(50)g mL^(-1)` since total amount of acid `= 5g` and volume of `H_(2)O = 100mL` and volume of `H_(2)O = 100mL` `:.` conc. of acid in `H_(2)O = ((5-a)/(100)) (g)/(mL)` Now, `K =("Conc. of acid in"C_(6)H_(6))/("Conc. of acid in"H_(2)O)` `= (a)/(50) xx (100)/((5-a))` `:. 4.1 =(a)/(50) xx (100)/((5-a))` or `a = 3.361g` `:.` Amount of acid in `50mL` `C_(6)H_(6) = 3.361 g` `:.` acid concentration in `C_(6)H_(6) = (3.361)/(50) xx 100 = 67.22 g//L` Also, amount of acid in `100mL H_(2)O = 5 -a = 5 - 3.361 = 1.639 g` `:.` acid concentration in `H_(2)O = (1.639)/(100) xx 1000 = 16.39g//L` |
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| 658. |
The vapour pressure of a solution containing non volatile solute is less than the vapour pressure of pure of solvent. Give reason. |
| Answer» In pure solvent the surface is oC Cupied by solvent molecules only. Number of molecules evpoarating will be more, hence vapour pressure is more. In case of solution the surface is oC Cupied by both solute and solvent molecules. The number of molecules evaporating will be less in case of solution containing non volatile solute. Hence the vapour pressure of solution containing non volatile solute is less than the vapour pressure of pure solvent. | |
| 659. |
How is molar mass related to the depression in freezing point of a solution ? |
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Answer» The expression for depression in Freezing point is `DeltaT_("f")=(K_("f")xx1000xx w)/("m"xx"W")` `K_("f")` = molal depression constant w = Weight of solute W = Weight of solvent m = molar mass of solute Molar mass of solute (m) `=("K"_("f")xx1000xx w)/(Delta"T"_("f")xx W)` `:.` Molar mass of solute (m) and depression in freezing point `(Delta"T"_("f"))` are inversely related. |
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| 660. |
` 15grams ` of methyl alcohol is dissolved in ` 35 grams `of water . What is the mass percentage of methyle alcohol in solution?A. ` 30% `B. ` 50% `C. ` 70% `D. ` 75% ` |
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Answer» Correct Answer - A Total mass of solution =`(15+35) g =50 `gram mass percentage of methyl alcohol `("Mass of methyl alcohol")/("Mass of solution")xx100` `(15)/(5)xx100= 30% ` |
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| 661. |
` 15grams ` of methyl alcohol is dissolved in ` 35 grams `of water . What is the mass percentage of methyle alcohol in solution?A. 0.3B. 0.5C. 0.7D. 0.75 |
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Answer» Correct Answer - A Total mass of solution`=(15+35)gram=50 "gram"` mass percentage of methyl alcohol`=("Mass of methyl alcohol")/("Mass of solution")xx100=15/50xx100=30%` |
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| 662. |
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. 488B. 32C. 64D. 128 |
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Answer» Correct Answer - C For dilute solutions, `(Deltap)/(p_(0))=(w_(B)xxm_(A))/(m_(B)xxw_(A))` `(185-183)/(185)=(1.2xx58)/(m_(B)xx100)` `m_(B)=64.38` `~~ 64 g mol^(-1)`] |
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| 663. |
Volume of `0.1M K_(2)Cr_(2)O_(7)` required to oxidise `35 mL` of `0.5M FeSO_(4)` solution isA. `29.2 mL`B. `17.5mL`C. `175 mL`D. `145 mL` |
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Answer» Correct Answer - A `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr 2Cr^(3+)+7H_(2)O` `Fe^(2+)rarr Fe^(3+)+e^(-)xx6` `Cr_(2)O_(7)^(2-)+6Fe^(2+)+14H^(+)rarr 2Cr^(3+)+7H_(2)O+6Fe^(3+)` `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))` `(Cr_(2)O_(7)^(2-)) ( Fe^(2+))` `(0.1xxV_(1))/(1)=(0.5xx35)/(6)` `V_(1)=(5xx35)/(6)=29.17 mL` |
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| 664. |
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")=0.5" 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. `" if X"_("benzene")=0.7" than Y"_("toluene")lt0.3` |
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Answer» Correct Answer - B From the graph it is clear that boiling point of pure benzene is lesser than that of pure toluene i.e. `"P"_("toluene")^(0)lt"P"_("benzene")^(0)("Benzene is more volatile than toluene")` `"X"_("benzene")lt"Y"_("benzene ")&" X"_("toluene")gt"Y"_("toluene")` |
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| 665. |
In isotonic solutions ________________.(i) solute and solvent both are same.(ii) osmotic pressure is same.(iii) solute and solvent may or may not be same.(iv) solute is always same solvent may be different. |
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Answer» (ii) osmotic pressure is same. (iii) solute and solvent may or may not be same. |
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| 666. |
Colligative properties are observed when……..A. a non-volatile solid is dissolved in a volatile liquidB. a non-volatile liquid is dissolved in another volatile liquidC. a gas is dissolved in non-voltatil liquidD. a volatile liquid is dissolved inanother volatial liquid |
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Answer» Correct Answer - A::B When only of one component of binary mixture either solvent or solute is volatile it causes deviation from ideal behavioru and vapour pressure of solution which causes change in colligative property. Hwnce (a) and (b) are correct. |
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| 667. |
Colligative properties are observed when _____________.(i) a non volatile solid is dissolved in a volatile liquid.(ii) a non volatile liquid is dissolved in another volatile liquid.(iii) a gas is dissolved in non volatile liquid.(iv) a volatile liquid is dissolved in another volatile liquid. |
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Answer» (i) a non volatile solid is dissolved in a volatile liquid. |
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| 668. |
Vapour pressure of `C_(6)H_(6)` and `C_(7)H_(8)` mixture at `50^(@)C` is given by `P(mm Hg) = 179 X_(B) +92`, where `X_(R)` is the mole fraction of `C_(6)H_(6)`. A solution is prepared by mixing 936g benzene and 736g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of `50^(@)C`. what would be mole fraction of `C_(6)H_(6)` in the vapour state? |
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Answer» Correct Answer - `0.9286` `P = 179 X_(B) +92` `P_(B)^(@) = 271, P_(T)^(@) = 92` `n_(B) = (936)/(78) = 12, n_(T) = (736)/(92) = 8` `X_(B) = (12)/(20) = 0.6 X_(T) = 0.4` `P_(T) = 271 xx 0.6 +92 xx 0.4 = 199.4` `Y_(B) = (271 xx 0.6)/(199.4) = 0.815` `Y_(T) = 0.185` On further condensation `X_(B) = 0.815, X_(T) = 0.185` `P_(T) = 271 xx 0.815 +92 xx 0.185 = 237.844` `Y_(B) = (271 xx 0.815)/(237.844) = 0.9286` |
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| 669. |
The vapour pressure of a certain liquid is given by the equation: `Log_(10)P = 3.54595 - (313.7)/(T) +1.40655 log_(10)T` where P is the vapour pressure in mm and `T =` Kelvin Temperature. Determine the molar latent heat of vaporisation as a function of temperature. calculate the its value at `80K`. |
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Answer» Correct Answer - `DeltaH = 1659.9` Cal. At `80 K, DeltaH = R[313.7 xx 2.303 +1.40655 T]` `(d InP)/(dT) = (DeltaH)/(RT^(2))` ...(i) `logP = 3.54595 - (313.7)/(T) +1.40655 logT` `In P = 3.54595 xx 2.303 - (313.7)/(T) xx 2.303 +1.40655 In T` `(d In P)/(dT) = (313.7 xx 2.303)/(T^(2)) +(1.40655)/(T)` ...(ii) Compairing equation (i) & (ii) `DeltaH = R[313.7 xx 2.303 +1.40655 T]` at `T = 80 K` `DeltaH = 1659.9 Cal`. |
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| 670. |
Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 torr. At the same temperature, a new solution of A and B having mole fractions `x_(A) and x_(B)`, respectively, has vapours pressure of 22. torr. The value of `x_(A)//x_(B)` in the new solution is ______. (Given that the vapour pressure of pure liquid A is 20 torr at temperature T). |
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Answer» For ideal solution `p=p_(A)^(2)x_(A)+p_(B)^(@)x_(B)` `45=P_(A)^(@)xx(1)/(2)+p_(B)^(@)xx(1)/(2)` `p_(A)^(@)+p_(B)^(@)=90` torr Given `p_(A)^(@)=20` torr, `:. p_(B)^(@)=70` torr For new solution : `p=p_(A)^(@)x_(A)+p_(B)^(@)x_(B)` `22.5=20x_(A)+70(1-x_(A))` `=70-50x_(A)` `x_(A)=(70-22.5)/(50)=0.95` `x_(B)=0.05` `:. (x_(A))/(x_(B))=(0.95)/(0.05)=19`. |
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| 671. |
A liquid is kept in a closed vessel . If a glass plate (negligible mass)with a small hole is kept on top of the liquid surface, then the vapour pressure of the liquid in the vesel is :A. More than what would be if the glass plate were removedB. Same as what would be if the glass plate were removedC. Less than what would be if the glass plate were removedD. Cannot be predicted |
| Answer» Correct Answer - B | |
| 672. |
For the dissolution of an ionic solid in waterA. hydration energy should be more than lattice energyB. lattice energy should be more than hydration energyC. lattice energy should be equal to hydration energyD. none of these |
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Answer» Correct Answer - D A level information |
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| 673. |
Why the boiling point of solution is higher than oure liquid? |
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Answer» Due to lowering in v.p. |
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| 674. |
A 1m solution can be more concentrated than a 1M solution ifA. density of solvent is more than 1B. density of solvent is equal to 1C. density of solvent is less than 1D. it is never possible |
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Answer» Correct Answer - A A level information |
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| 675. |
Which of the following statements is false ?A. Units of atmospheric pressure and osmotic pressure are the sameB. In reverse osmosis, solvent molecules move through a semipermeable memreane from a region of lower concentration of solute to a region of higher concentration.C. The value of molal depression constant depends on nature of solvent.D. Relative lowering of vapour pressure is a dimensionless quantity. |
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Answer» Correct Answer - b is the false statement in reverse osmosis, solvent molecules move through a semipermeable membrane from a reglon of higher solute concentration to a region of lower solute concentration. |
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| 676. |
Consider the Fig. and make the correct option.A. Water will move form side (A) to side (B) if a pressure lower than osmotic pressure is applied on poston (B).B. water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piton (B).C. water will move from side (B) to side (A) if a pressure equal too osmotic pressure is applied on piston (B).D. water will move from side (A) to dide (B) if pressure equal to osmotic pressure is applied on piston (A). |
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Answer» Correct Answer - b is the correct option. This is because of revarse osmosis which is taking place. |
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| 677. |
Out of 1M and 1m aqueous solution which is more concentrated? |
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Answer» 1M as density of water is 1 gm/Ml |
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| 678. |
Calculate the volume of 75% of H2SO4 by weight (d=1.8 gm/ml) required to prepare 1L of 0.2 M solution. |
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Answer» M1 = P x d x 10 /98 M1 V1 = M2V2 = 14.5 ml |
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| 679. |
Which statement best explains the meaning of the phases "like dissolve like"?A. A Solute will easily dissolve a solute of similar massB. A solvent and solute with similar intermolecular forces will readily form a solutionC. The only true solutions are formed when water dissolves a non-polar soluteD. The only true solutions are formed when water dissolves a polar solute |
| Answer» Correct Answer - B | |
| 680. |
Henry law constant for two gases are 21.5 and 49.5 atm ,which gas is more soluble. |
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Answer» KH is inversely proportional to solubility. |
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| 681. |
The value of Henry constant `K_(H)` isA. greater for gases with higher solubilit.B. grater for gases with lower solubility.C. constanat for all gases.D. not related to the solubility of gases. |
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Answer» Correct Answer - b is the correct answer. In general, higher the value of `K_(H)` lower is the solubility of gas in a liquid. |
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| 682. |
Name the law which explains the relationship between solubility of a gas in a liquid and pressure above the liquid surface. Also write the name of the variable which is kept constant for this law. |
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Answer» This is Henry’s law and temperature is kept constant for this law. [According to Henry's law the partial pressure of gas in vapour phase is proportional to the mole fraction of gas in the solution at constant temperature.] |
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| 683. |
Assertion:Osmotic pressure of non-aqueous solution can be determine by Berkeley-Hartley method. Reason: the semipermeable membrance used in Berkeley-Hartley method is `Cu_(2)[Fe(CN)_(6)].`A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both the assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - D `Cu_(2)[Fe(CN)_(6)]`is soluble in non-aqueous solution. |
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| 684. |
The addition of `0.643 g` of a compound to `50 mL` of benzene (density 0.879 g `mL^(-1)`) lowers the freezing point from `5.51` to `5.03^(@)C`. If `K_(f)` for benzene is `5.12`, calculate the molecular weight of the compound. |
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Answer» Correct Answer - `Mw=156.06 g mol^(-1)` According to depression in freezing point method `Mw_(2)=(1000xxK_(f))/(DeltaT_(f)) xx W_(2)/W_(1)` Given that `K_(f) rarr` molar depression constant of benzene `=5-12 K kg "mol"^(-1)` `W_(2)` =Weight of compound =0.643 g `W_(1)` =Weight of benzene =volume`xx` Density `=50 xx 0.879` `=43.95 g` `DeltaT_(f) rarr` Depression in freezing point =`5.51-5.03=0.48^(@)C` Therefore, on substitution, `Mw_(2)=(1000xx5.12xx0.643)/(0.48xx43.95)=156.06` |
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| 685. |
Addition of `0.40 g` of a compound to `45.5 m L` of benzene (density `0.879 g mL^(-1)`) lowers the freezing point from `5.51^(@)C` to `4.998^(@)C`.If `K_(f)` for benzene is `5.12 K kg mol^(-1)`,calculate the molar mass of the compound. |
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Answer» The given values are: `W_("solute")=0.40 g`,`K_(f)=5.12 K m^(-1)` `rho_("benzene")=0.879 g mL^(-1)` `V_("benzene")=45.50 mL` `DeltaT_(f)=T_(f)^(@)-T_(f)=5.51-5.03=0.212^(@)C` Now weight of benzene =`V xx rho` `=45.50xx0.879` `=40.00 g` Molar mass of solute,`M_("solute")` is calculated as `Mw_("solute")=(K_(f)xxW_("solute")xx1000)/(W_("solvent")xxDeltaT_(f))` `=(5.12xx0.40xx1000)/(40xx0.512)=400` Thus, the molecular weight of solute is `400 g mol^(-1)`. |
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| 686. |
How does Henry constant vary with the solubility of gas in the solution at a given pressure? |
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Answer» Higher the value of “Kh” (Henry constant), lower is the solubility of gas in solution. |
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| 687. |
State how does osmotic pressure vary with temperature? |
| Answer» Osomotic pressure increases with increase in temperature. | |
| 688. |
Which are the correct statementsA. `CHCl_(3)+CCl_(4)` - Endothermic solutionB. Acetic acid +Pyridine - Hot solutionC. `HNO_(3)+Water - Endothermic solutionD. Water+HI - Minimum azeotrope |
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Answer» Correct Answer - 1,2 |
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| 689. |
When dehydrated fruits and vegetables are placed in water, they slowly swell and return to original form. Why? Would a temperature increase accelerate the process? Explain. |
| Answer» The cell walls of fruits and vegetables are semi-permeable. The liquid inside the cells in the dried fruits and vegetables is more concentrated. When these dried fruits and vegetables due to osmosis and they swell and return to their original form. Since the increase in temperature increases the osmotic pressure `(phi = T)` , the process gets accelerated. | |
| 690. |
V litre decinormal solution of NaCl is prepared. Half of the solution is converted into centionormal and added to the left decinormal solution. ThenA. Number of millimoles of NaCl are reduced by `(1)/(5)`B. Number of milliequivalents of NaCl do not changeC. Normality of the final solution becomes 0.01 ND. Molarity of the final solution becomes 0.018 M |
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Answer» Correct Answer - 2,4 |
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| 691. |
Outer hard shells of two eggs are removed. One of the eggs is placed in pure water and the other is placed in saturated solution of sodium chloride. What will be observed and why? |
| Answer» The egg placed in water will swell due to the osmosis of pure water into the egg. On the other hand, the egg placed in the saturated solution of `NaCl` will shrink due to the osmosis of water out of the egg. This is because osmosis always occurs from higher concentration of solvent to lower concentration of solvent. | |
| 692. |
After removing the hard shell of an egg by dissolving in dilute`HCl`a semipermeable membrance can be visible .If such an egg is kept in a saturated solution of common salt, the size of egg willA. ShrinkB. GrowC. Remain unchangedD. First shrink ,then grow |
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Answer» Correct Answer - A Exosmosis will take place |
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| 693. |
75 g ethylene glycol is dissolved in 500 gram water. The solution is placed in a refrigerator maintained at a temperature of 263.7 K. What amount of ice will separate out at this temperature ? (`K_(f)" water"=1.86 K "mollality"^(-1)`)A. 300 gB. 200 gC. 178 gD. 258 g |
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Answer» Correct Answer - 4 |
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| 694. |
At 300 K, the vapour pressure of an ideal solution containing 1 mole of A and 3 moles of B is 500 mm Hg. At the same temperature, 2 moles of B are added to this solution. The vapour pressure of solution increases by 10% of the original vapour pressure. Correct statements about the vapour pressure areA. Vapour pressure of A in the pure state is 50 mm HgB. Vapour pressure of B in the pure state is 650 mm HgC. Ratio of final pressure to the initial vapour pressure is 1:0.5D. Ratio of vapour pressure of pure B to the vapour pressure of pure A is `13:1` |
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Answer» Correct Answer - 1,2,4 |
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| 695. |
Consider the following arrangement and choose the correct options A. O.P. of `NaSO_(4)` solution is lesser than the O.P. of KCl solutionB. Water will flow from KCl solution to `Na_(2)SO_(4)` solutionC. Water will flow `Na_(2)SO_(4)` solution to KCl solutionD. O.P. of `Na_(2)SO_(4)` solution is higher than the O.P. of KCl solution |
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Answer» Correct Answer - 2,4 |
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| 696. |
When 50mL of ethanol and 50mL of water are mixed, predict whether the volume of the solution is equal to, greater than or less than 100 mL. |
| Answer» In both ethanol and water, the molecules are hydrogen bonded. When these are mized to form the solution, the molecules of one liquid will tend to break the hydrogen bonds in the molecules of the other liquid and vice versa. The attractive forces in the molecules decrese and this leads to increase in volume `(DeltaV_(mixing) is +ve). Therefore, the volume of the solution is more than 100 mL. | |
| 697. |
On the basis of concept of lattice enegy and hydration energy explain the fact that NaCI is soluble in water but not in banzene. |
| Answer» In aqueous solution, the magnitude of hydration energy (energy released) for sodium shoride is more than that of lattice energy (energy required). Therefore, the salt is soluble in water. In benzen solvent which is of non-polar nature reverse happens. Therefore, salt does not dissolve in benzene. | |
| 698. |
Outer hard shell of two eggs are removed by placing in dilute HCI. If one is kept in pure water and the other in soldium chloride (NaCI) solution, what will you notice. |
| Answer» The outer hard shell consists of calsistes of calcium carbonate. It dissolves in dilute HCI. The soft layer which is left actes as a semipermeable membreane. The egg placed in water will swell in size because water (hypotonic in nature) will enter the semiermeable membrane of the egg due to osmosis. However, in the second case, fluid from the egg (hypotonic) will come out and will pass into and will pass into sodium chloride solution which is at a higher osmotic pressure (hypertonic). Therefore, the egg placed in sodium chloride sodium chloeide solution or saline water will shring in size. | |
| 699. |
The freezing point of a solution containing 2.40 g of a compound in 60.0g of benzene is `0.10^(@)"C"` lower than that of pure benzene. What is the molecular weight of the compound?`("K"_("f")" is " 5.12^(@)"C"//"m for benzene"` |
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Answer» Correct Answer - `"2048 g/mol"` Solute-B ,Solvent-A `Delta"T"_("f")=("K"_("f")xx"w"_("B")xx1000)/("m"_("B")xx"w"_("A"))` `0.10=(5.12xx2.40xx1000)/("m"_(B")xx60)` `"m"_("B")=2048"gm mol"^(-1)` |
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| 700. |
Assiertion : The molecular mass of acetic acid was determined by depression in benzene and was found to be different than in water. Reason : Water is polar and benzene in non-polar.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 - a Reason is the correct explanation for assertion. |
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