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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.
| 1751. |
Which of the following expressions is correct?A. `N=(meq)/L`B. `N=(meq)/(mL)`C. `N=(mmol)/("molar mass")`D. `N=(meq)/(mg//meq)` |
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Answer» Correct Answer - 2 Normality of a solution is defined as the number of equivalents of solute per litre of solution. By definition, there are 1000 milliequivalents (meq) in one equivalent of so,ute and 1000 milliliters in one litre of solution. Thus, normality may also be repersented as `Normality=("Number of milliequivalents of solute")/("Millilitres of solution")` `=(no . meq)/(mL)` |
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| 1752. |
A sugar syrup of weight 183.42 g contains 3.42 g of sugar `(C_(12)H_(22)O_(11))`. Calulate the mole fraction of sugar. |
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Answer» Correct Answer - 0.00099 `"Mass of sugar" = 3.42 g: "Moles of sugar" =((3.42g))/((342 g mol^(-1)))=0.01 mol` `"Mass of water"=183.42-3.42=180g : "Moles of water"=(180g)/((342 "g mol"^(-1)))=10 mol` `"Mole fration of sugar" = ("No. of moles of sugar")/("No. of moles of sugar + No. of moles of water")` `=((0.01mol))/((0.01mol)+(10 mol))=(0.01)/(10.01)=0.00099`. |
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| 1753. |
The mole fraction of sugar `(C_(12)H_(22)O_(11))` in an aqueous solution that is `5.30%` sugar by mass isA. 0.99706B. 0.00294C. 0.79906D. 0.20094 |
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Answer» Correct Answer - 2 To calculate the mole fraction, we must find the number of moles of sugar and of water in some fixed quantity of solution. From the given mass percent, we can say that there is `5.30 g` of sugar in every `100 f` of solution. Since the solution contains only sugar and water, `100 g` of solution must contain `100 g-5.30 g=94.70 g` of water. A table of atomic masses shows that the molecular mass of the water is `18 u` and the molecular mass of sugar is `342 u`. Therefore, in the `100 g` sample of solution there is: `n_(sugar)=("mass of sugar")/("molar mass of sugar")` `=(5.30 g sugar)/(342 g sugar//"mol sugar")` `=0.0155` mol sugar `n_(water)=("mass of water")/("molar mass of water")` `=(94.70" g water")/(18" g sugar"//"mol water")` `=5.26 mol H_(2)O` We can substitute thses date into Equations `(2.9 and 2.10)`, which give the mole fraction relationship for a two-component solution: `chi_(C_(12)H_(22)O_(11))=n_(C_(12)H_(22)C_(11))/(n_(C_(12)H_(22)C_(11))+n_(H_(2)O))` `=(0.0155 mol sugar)/(0.0155 mol sugar + 5.26 mol H_(2)O)` `=0.00294` Since `chi_(C_(12)H_(22)O_(11))+chi_(H_(2)O)=1` `chi_(H_(2)O)=1-chi_(C_(12)H_(22)O_(11))` `=1-0.00294` `=0.99706` |
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| 1754. |
An aqueous solution containing `28%` by mass of liquid A `(mol.mass = 140)` has a vapour pressure of `160 mm` at `30^@C`. Find the vapour pressure of the pure liquid A. (The vapour pressure of the water at `30^@C` is `150 mm`.) |
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Answer» For two miscible liquids , `P_("total")`= Mole fraction of `A xx P_(A)^(@)` `+`Mole fraction of `B xx P_(B)^(@)` Number of moles of` A = 28/140=2` Liquid B is water. Its mass is `(100-28)`,i.e., `72`. Number of moles of `B=72/18 = 4.0` Total number of moles `=0.2 + 4.0 =4.2` Given ,`p_("total") = 160 mm` `p_(B)^(@) = 150 mm` So, `160=0.2/4.2 xx p_(A)^(@) + 4.0 /4.2 xx 150` ` p_(A)^(@)=(17.15 xx 4.2)/0.2 = 360.15 mm` |
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| 1755. |
The vapour pressure of a certain pure liquid A at 298 K is 40 mbar. When a solution of B is prepared in A at the same temperature, the vapour pressure is found to be 32 mbar. The mole fraction of A in the solution isA. `a.) 0.5`B. `b.)0.3`C. `c.)0.4`D. `d.)0.8` |
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Answer» Correct Answer - d d. `p=p^(@)chi_(A)` `32=40 xx chi_(A) or chi_(A) = 0.8` |
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| 1756. |
`100 mL` of liquid A and `25 mL` of liquid B are mixed to form a solution of volume `125 mL`. Then the solution isA. a.) IdealB. b.) Non-ideal with positive deviationC. c.) Non-ideal with negative deviationD. d.) Cannot be predicted |
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Answer» Correct Answer - a `Delta_(mix)V=0`, hence the solution is ideal. |
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| 1757. |
What volume of diethly ether must be mixed with `0.125 L` water to form a solution in which the volume precent of the diethyl ether is `3.2%`?A. `0.0041 L`B. `0.0037 L`C. `0.0027 L`D. `0.0078 L` |
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Answer» Correct Answer - 1 Volume -Volume units are convenient for liquid-liquid solutions. Volume rather than the masses of the components. It is defined by Volume precent `=("Volume of pure component")/("sun of the volumes of each component")xx100%` For a two components solution: `"Volume precent"_(1)=V_(1)/(V_(1)+V_(2))xx100%` where `V_(1)` is the volume of component when it is pure and `V_(2)` is the volume of component when it is pure. In general, the volume of the solution is not sum of the volumes of the individual components. For example, a mixture of `0.500 L` water and `0.500 L` ethanol occupies a volume of only `0.965 L`. In the present case, let x equal the required volume (in litres) of diethyl ether. Then the volume precent of diethyl ether is defined as `x/(x+0.125 L)xx100%=3.2%` `x=0.0041 L` diethyl ether |
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| 1758. |
Assertion (A): The dissolution of gases in water is always an endothermic process. Reason (R ) : The dissolution of gases in water proceed with a negative value of `DeltaS`.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is correct, but `(R )` is correct. |
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Answer» Correct Answer - D The dissolution of gasses in water requires energy, hence it is endothermic in nature, but during dissolution entropy may increase or decrease. |
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| 1759. |
Assertion (A): Water boiling at `100^(@)C` at `1` atmospheric pressure in a beaker is not at equilibrium. Reason (R ): If refers to an open system.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is correct, but `(R )` is correct. |
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Answer» Correct Answer - A In an open system, equilibrium can never be achieved. |
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| 1760. |
`100 mL` of liquid A and `25 mL` of liquid B are mixed to form a solution of volume `125 mL`. Then the solution isA. idealB. non-ideal with positive deviationC. non-ideal with negative deviationD. cannot be predicted |
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Answer» Correct Answer - A `DeltaV_("mix")=0`, hence the solution is ideal.] |
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| 1761. |
The vapour pressure of pure benzene at `88^(@)C` is `957 mm` and that of toluene at the same temperature is `379.5 mm`. The composition of benzene-toluene mixture boiling at `88^(@)C` will beA. `x_("benzene")=0.66,x_("toluene")=0.34`B. `x_("benzene")=0.34,x_("toluene")=0.66`C. `x_("benzene")=x_("tolulene")=0.5`D. `x_("benzene")=0.75,x_("toluene")=0.25` |
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Answer» Correct Answer - A `p=p_("benzene")^(0)x_("benzene")+p_("tolulene")^(0)x_("toluene")` `760=957.x_("benzene")+379.5(1-x_("benzene"))` `x_("benzene")=0.66` and`x_("toluene")=1-0.66=0.34`] |
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| 1762. |
1000 gram aqueous solution of `CaCO_(3)` contains 10 gram of carbonate. Concentration of solution is:A. 10 p pmB. 100 ppmC. 1000 ppmD. 10,000 ppm |
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Answer» Correct Answer - D `10^(3)`gm solution contains`=10 gm` of carbonate So in `10^(6) gm =(10)/(10^(3)) xx 10^(6) = 10^(4)` |
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| 1763. |
What does normality (N) represent ? |
| Answer» It represents the number of gram equivalents of the solute dissolved per litre of the solution. Its unit is equiv. `L^(-1)`. | |
| 1764. |
How can you compere the relative solubilities of different slats in the same solvet ? |
| Answer» The relative solubilies can be compared in terms of solubility product constant `K_(sp)` values i.e. more the `K_(sp)` value, more will be the solubility of the sakt. | |
| 1765. |
Give one example of interstitial of interstitial and substitutional solid solutins. |
| Answer» Tungseten carbide is an example of interstitial solid solution while brass (allo of Cu and Zn) represents substitutional solid solution. | |
| 1766. |
Mole fraction of component `A` in vapour phase is `chi_(1)` and that of component `A` in liquid mixture is `chi_2`, then (`p_(A)^@`)= vapour pressure of pure A, `p_(B)^@` = vapour pressure of pure B), the total vapour pressure of liquid mixture isA. `(p_(A)^(@)chi_(2))/chi_(1)`B. `(p_(A)^(@)chi_(1))/chi_(2)`C. `(p_(B)^(@)chi_(1))/chi_(2)`D. `(p_(B)^(@)chi_(2))/chi_(1)` |
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Answer» Correct Answer - A `chi_(A)^(V)=(P_(A)^(@)chi_(A))/(P_("total"))=chi_(1)(P_(A)^(@)chi_(2))/(P_("total")) ` or `P_("total")=P_(A)^(@)(chi_(2)/chi_(1))` |
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| 1767. |
If `P^(@)` and `P_(s)` are vapour pressure of solvent and its solution, respectively, `chi_(1)` and `chi_(2)` are mole fractions of solvent and solute, respectively, thenA. `P_(s)=P^(@)//chi_(2)`B. `P^(@)-P_(s)=P^(@)chi_(2)`C. `P_(s)=P^(@)chi_(2)`D. `(P^(@)-P_(s))/P_(s)=chi_(1)/(chi_(1) + chi_(2))` |
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Answer» Correct Answer - B `(P^(@)-P_(S))/P^(@)` = Mole fraction of solute = `chi_(2)` |
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| 1768. |
Haven’t you noticed the caption “Shake well before use” in certain medicine bottles?a. To which class do the substances in them belong to? (Colloid, Solution, Suspension)b. What is the relevance of such instructions on these bottles? |
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Answer» a. Suspension b. It is used after shaking well because it is settled down after a long time. |
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| 1769. |
If `P^(@)` and `P_(s)` are vapour pressure of solvent and its solution, respectively, `chi_(1)` and `chi_(2)` are mole fractions of solvent and solute, respectively, thenA. `P=P_(0)N_(2)`B. `P=P_(0)N_(1)`C. `P_(0)=PN_(1)`D. `P=P_(0)(N_(1)//N_(2))` |
| Answer» Correct Answer - B | |
| 1770. |
Assertion: The concentration of pollutants in water or atmosphere is often expressed in terms of ppm Reason :Concentration in parts per million can be expressed as mass to mass, volume of volume and mass to volume .A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - B When a solute is present in trace quantities it is convenient to express concentration in ppm |
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| 1771. |
Given below are some solutions. Identify and write down the solvent and the solute present in them. (Hint: Those present in large amount is the solvent and that in small amount is the solute).SolutionSolventSoluteSalt waterOrnamental goldSoda waterDilute hydrochloric acid |
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| 1772. |
Assertion:Amalgam of mercury with sodium is an example of solid solutions . Reason:Mercury is solvent and sodium is solute in the solutionA. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - C Amalgam of mercury with sodium is an example of liquid in solid type solid solution. Here mercury (liquid metal) acts as solute and sodium as solvent. |
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| 1773. |
In amalgam of mercury with sodium, solvent isA. mercuryB. sodiumC. amalgamD. none of these |
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Answer» Correct Answer - B Amalgam of mercury with sodium is an example of solid solution in which mercury (liquid ) is solute and sodium (solid ) is a solvent. |
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| 1774. |
`18` carat gold contains `18` parts in `24` parts and has a fineness ofA. `0.75B. `7.5C. `75D. `750 |
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Answer» Correct Answer - 4 Fineness of gold is the quantity of gold in an alloy expressed as parts per thousand: `18/24xx1000=750` Chemical similarity and nearness of molecular sizes two factors which play a very important role in forming solid solution. |
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| 1775. |
The vapour pressure of benzene at a certain temperature is `640 mm` of Hg A non-volatile and non-electrolyte solid weighing `0.175g` is added to `39.08g` of benzene .the vapour pressure of the solution is`600 mm` of Hg.What is the molecular weight of solid substance ?A. `49.50`B. `59.6`C. `69.5`D. `79.8` |
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Answer» Correct Answer - C `(P_(0)-P_(s))/(P_(0))=((w)/(m))/((w)/(m)+(W)/(M)) because (W)/(M)gt(w)/(m)` `rArr(640-600)/(640)` `=(w)/(m)xx(M)/(W)rArr(40)/(640)=(2.175xx78)/(mxx39.08)` `m=(2.175xx78)/(39.08)xx(640)/(40)` `m=69.45` |
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| 1776. |
Explain the effect of temperature on the vapour pressure of a liquid. |
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Answer» The vapour pressure of a liquid is the pressure of the vapour in equilibrium with the liquid at a given temperature. The evaporation of a liquid requires thermal energy. Hence, as temperature rises, the vapour pressure rises until it becomes equal to the external pressure, generally the atmospheric pressure, 101.3 kNm-2 (1 atm). This temperature is called the normal boiling point of the liquid. |
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| 1777. |
A compound `MX_(2)`has observed and normal molar masses `65.6 and 164`respectively .Calculated the apparent degree of ionization of `MX_(2)`:A. `75%`B. `85%`C. `65%`D. `25%` |
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Answer» Correct Answer - A `i=("normal molar mass")/("Observed molar mass")` `=(164)/(65.6)=2.5` `alpha=(i-1)/(n-1),n=3`(number of ions) `(2.5-1)/(3-1)=(1.5)/(2)=0.75` `:.` percentage ionization of `MX_(2)`will be `75%` |
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| 1778. |
Calculate the percentage degree of dissociation of an electrolyte `XY_(2)` (Normal molar mass `=164`) in water if the observed molar mass by measuring elevation in boiling point is `65.6`: |
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Answer» Correct Answer - `75%` `{:(,XY_(2),hArr,X^(2+),+,2Y^(-)),("Initially",1,,0,,0),("at equilibrium",1-alpha,,alpha,,2alpha):}` Total no. of moles `=1-alpha+alpha+2alpha=1+2alpha` `i=("Normal molar mass")/("Observed molar mass"), (1+2alpha)/(1)=(164)/(65.6)` `therefore alpha=0.75,%alpha=75%` |
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| 1779. |
A complex of iron and cyanide ions is `100%`ionised ar `1`m (molal). If its elevation in b.p.is `2.08` then the complex is `(K_(b)=0.52^(@)mol^(-1)kg)`,A. `K_(3)[Fe(CN)_(6)]`B. `Fe(CN)_(2)`C. `K_(4)[Fe(CN)_(6)]`D. `Fe(CN)_(4)` |
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Answer» Correct Answer - A `DeltaT_(b)=iK_(b)m` so `i=(2.08)/(0.52xx1)=4` so the compex is `K_(3)[Fe(CN)_(6)]` `K_(3)[Fe(CN)_(6)]hArr3K^(+)+[Fe(CN)_(6)]^(3-)` |
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| 1780. |
The vapour pressure of a dilute aqueous solution of glucose is 740 mm of mercury at 373 K. The mole fraction of the solute is:A. `(1)/(20)`B. `(1)/(38)`C. `(1)/(76)`D. `(1)/(760)` |
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Answer» Correct Answer - B `(p^(@)-p_(s))/(p^(@)) = x_(B)` (mole fraction of the solute) At `373 K (100^(@)C) V.P.` of pure water, `p^(@) = 760`mm of Hg `:.` mole fraction of solute `= (760 - 740)/(760) = (1)/(38)` |
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| 1781. |
The vapour pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373 K . The mole fraction of solute is :A. `(1)/(10)`B. `(1)/(76)`C. `(1)/(35)`D. `(1)/(76)` |
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Answer» Correct Answer - D `(p-p_(2))/(p)=` Mole fraction of solute p = Vapour pressure of water at 373 K is 760 mm] |
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| 1782. |
If `5.85`g of `NaCl`are dissolved in`90g` of water,the moles fraction of`NaCl` isA. `0.1`B. `0.2`C. `0.3`D. `0.0196` |
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Answer» Correct Answer - D `5.85gNaCl=(5.85)/(58.5)"mole" =0.1mol` `90 gH_(2)O=(90)/(18)"moles" =5"moles"` Mole fraction of `NaCl =(0.1)/(5+0.1)~~0.0196` |
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| 1783. |
The osmotic pressure of blood is 7.40 atm at `27^(@) C`.The number of mol of glucose to be used per litre for an intravenous injection that is to have the same osmotic pressure as blood isA. 0.3B. 0.2C. 0.1D. 0.4 |
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Answer» Correct Answer - A `pi=n/V(RT)` `n=(piV)/(RT) =(7.40 xx 1)/(0.0821 xx 300) =0.3` |
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| 1784. |
A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature ?(a) 5.08 atm (b) 2.54 atm (c) 4.92 atm (d) 2.46 atm |
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Answer» Correct answer is (c) 4.92 atm |
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| 1785. |
A 5% solution of glucose is isotonic with 1.1 % solution of KCl at `30^(@)C`. Calculate the degree of ionisation of KCl. |
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Answer» Correct Answer - 0.88 or 88% |
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| 1786. |
The osmotic pressure of blood is `8.21 atm` at `310 K`. How much glucose should be used per L for an intravenous injection that is isotopic with blood? |
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Answer» The given values are `pi_("blood")`=8.21 atm , `V= 1 L`,`T=310 K` `pi_("blood")=8.21 atm` (blood and glucose are isotonic) Now, `piV=nRT` or `n=(piV)/(RT)` `:.n=(8.21xx1.0)/(0.0821xx300)=10/31` `:.` Weight of glucose `=nxx` Molecular weight `=(10)/(31)xx180=58.06 g` |
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| 1787. |
Why naphthalene dissolves in benzene but not in water ? |
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Answer» Since naphthalene is a covalent nonpolar substance, it is soluble in a nonpolar solvent like benzene but insoluble in polar solvent like water. |
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| 1788. |
A living cell contains a solution which is isotonic with `0.3`(M) sugar solution .What osmotic pressure develops when the cell is placed in `0.1(M)KCl` solution at body temperature?A. 5.08 atmB. 2.54 atmC. 4.92 atmD. 2.46 atm |
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Answer» Correct Answer - B `pi=(i_(1)C_(1)-i_(2)C_(2))RT=(1xx0.3xx-2xx0.1)xx0.082xx310` `2.54` atm. |
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| 1789. |
A temperature at which 0.1 M KCl solution will have osmotic pressure 10 atm will be :(a) 408 °C (b) 263 °C (c) 310 °C (d) 337 °C |
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Answer» Option : (d) 337 °C |
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| 1790. |
The osmotic pressure of 0.2 M KCl solution at 310 K is :(a) 10.17 atm (b) 5.084 atm (c) 8.36 atm (d) 12.2 atm |
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Answer» Option : (a) 10.17 atm |
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| 1791. |
A 0.2 m aqueous solution of KCl freezes at -0.68°C. Calculate ‘i’ and the osmotic pressure at 0°C. Assume the volume of solution to be that of pure H2O and Kf for H2O is 1.86 K Kg/mol. |
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Answer» (ΔTf)normal = Kf x m = 1.86 x 0.2= 0.372 i= Observed colligative property/Normal colligative property = 0.68/0.372 Observed osmotic pressure= i x Normal osmotic pressure = i x c RT = 1.83 x 0.2 x 0.082 x 273 = 8.2 atm |
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| 1792. |
Assertion: 1 M solution of KCl has greater osmotic pressure than 1 M solution of glucose at the same temperature Reason :In solution KCl dissociates to produce more number of particles .A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false |
| Answer» Correct Answer - A | |
| 1793. |
Assertion (A): `0.1 M` solution of `NaCl` has greater osmotic pressure than `0.1 M`solution of glucose at same temperature. Reason (R ): In solution, `NaCl` dissociates to produce more number of particles.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is correct, but `(R )` is correct. |
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Answer» Correct Answer - A For `NaCl`,`i=2`, while for glucose, i=1.Hence, osmotic pressure of `NaCl` is higher than that of glucose solution. |
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| 1794. |
Assertion (A): Larger the value of cryoscopic constant of the solvent, lesser will be the freezing point of the solution. Reason (R ): Depression in the freezing point depends on the nature of the solvent.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is correct, but `(R )` is correct. |
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Answer» Correct Answer - A Depression in freezing point is a colligative property, hence its value depends upon the number of solute particles. |
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| 1795. |
What is the percent by mass of iodine needed to reduce the freezing point of benzene to `3.5^(@)"C"`? The freezing point and cryoscopic constant of pure benzene are `5.5^(@)"C"` and `5.12K/m` respectively. |
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Answer» `Delta"T"_(f)="T"_(f)^(0)-"T"_(f)="K"_(f)."m"` `5.5^(@)"C"-3.5^(@)"C"=5012xxm` `m=2/5012=0.39" molal"` `therefore ` Mass of iodine needed for 1000g of benze`=mxx"molecular mass of iodine "I^(2)` `=0.39"mol//kgxx254g//mol=99.06g//kg` `therefore` 1000g+99.06g solution contain `99.06g "I"^(2)` 100g solution contains `99.06gxx100/1099.06g=9.01%` |
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| 1796. |
The cryoscopic constant for acetic acid is 3.6 K kg/mol. A solution of 1g of a hydrocarbon in 100 g of acetic acid freeze at `16.14^(@)"C"` instead of the usual `16.60^(@)"C"`. The hydrocarbon contains 92.3% carbon. What is the molecular formula? |
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Answer» Correct Answer - `C_(6)H_(6)` Hydrocrabon-B ,acetic acid-A `"m"_("B")=(1000"K"_("f")"w"_("B"))/("w"_("A")xxDelta"T"_("f"))` `=(1000xx3.6xx1)/(100xx(16.60-16.14))` `"m"_("B")=78 "gm"` let molecular formula of Hydrocarbon`="C"_("x")"H"_("y")` `12"x"+"y"=78` `implies(12"x")/(78)xx100=92` `implies" x"=(92xx78)/(12xx100)` `"x=6(App)"` `"y=6"` `therefore"Molecular formula of compound"-"C"_(6)"H"_(6)` |
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| 1797. |
At the higher altitudes the boiling point of water lowers becauseA. Atmospheric pressure is lowB. Temperature is lowC. Atmospheric pressure is highD. none of these |
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Answer» Correct Answer - A The boiling occurs at lowers temperature if atmospheric pressure is lower than`76`cm Hg. |
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| 1798. |
Cryoscopic constant of a liquid is:A. decreases in freezing point when 1 gram of solute is dissolved per kg of the solventB. decreases in the dreezing point when 1 mole of solute is dissolved per kg of the solventC. the elevation for 1 molar solutionD. a factor used for calculation of elevation in boiling point |
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Answer» Correct Answer - B `DeltaT_(f) = K_(f), m` If `m = 1, DeltaT_(f) = K_(f)` |
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| 1799. |
The relative lowering in vapour pressure is:A. `prop X_("solute")`B. `prop (1)/(X_("solute"))`C. `=X_("solute")`D. `prop m` |
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Answer» Correct Answer - C `(P^(@)-P_(s))/(P^(@)) = X_("solute")` |
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| 1800. |
A `5.8%` solution of `NaCI` has vapour pressure closets to:A. 5.8% solution of ureaB. 2m solution of glucoseC. 1m solution of ureaD. 5.8% solution of glucose |
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Answer» Correct Answer - B 5.8% solution of `NaCI` is approximately 1 molal solution. As `NaCI` dissociates in `Na^(+)` and `CI^(-)` the concentration of particles is approx 2m. |
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