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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.
| 1301. |
`CNS^(ө)` ions give red colour with `Fe^(3+)` ions in aqueous solution as: `Fe^(3+)(aq) + 3CNS^(ө)(aq) rarr Feunderset("red")((CNS))_(3)(aq) ` If `0.1 M` `KCNS` solution is separated from `0.1 M FeCl_(3)` solution by means of a semi-permeable membrane, red colour will appear on: a.`FeCl_(3)` soluion , b. `KCNS` solution side , c. Both sides , d. Neither side |
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Answer» d. `Fe^(3+)(aq) + 3CNS^(ө)(aq) rarr Feunderset("Red")((CNS))_(3)(aq)` `rArr` Only solvent molecules moves from dilute solution to concentrated solution. `{:(underset("(Dilute)")(underset(C_("effective")=0.2 M)(0.1MKCNS)),,,underset(C_("Concentrated"))underset(C_("effective")=0.2M)(0.1MFeCl_(3))):}` So , no red color will appear. |
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| 1302. |
At `17^(@)C`, the osmotic pressure of sugar solution is 580 torr. The solution is diluted and the temperature is raised to `57^(@)C`, when the osmotic pressure is found to be 165 torr. The extent of dilution is a.2 times ,b.3 times ,c.4 times ,d.5 times |
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Answer» c.Given `T_(i)=17^(@)C=290 K,T_(f)=57^(@)C=330 K` `pi(i)=580 "torr"=580/760 mm Hg` `pi(f)=165 "torr"=165/760 mm Hg` Using equation `pi=CRT` `580/760=C_(i) xx R xx 290` ….(i) `165/760=C_(f) xx R xx 330` ….(ii) Dividing Eq. (i) by Eq. (ii), we get `C_(i)/C_(f)=2/1//2=4 rArr` Dilution = 4 times |
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| 1303. |
What is Henry’s Law? |
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Answer» The solubility of a gas at a given temperature is directly proportional to the pressure at which it is dissolved. P=KH X If KH increases solubility decreases. |
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| 1304. |
At a constant temperature , `DeltaS` will be maximum for which of the following processes:A. Vaporisation of a pure solventB. Vaporisation of solvent from a solution containing nonvolatile and nonelectrolytic solute in itC. Vaporisation of solvent from a solution containing nonvolatile but electrolytic solute in itD. Entropy change will be same in all the above cases |
| Answer» Correct Answer - A | |
| 1305. |
Calculate the proportion of O2 and N2 dissolved in water at 298 K. When air containing 20% O2 and 80% N2 by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O2) = 4.6 x atm and KH(N2) 8.5 x 104 atm. |
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Answer» C1V1 = C2V2 6M(V1) = 0.25M x 500 ml V1 = \(\frac{0.25\times500}{6}\) V1 = 20.3 mL |
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| 1306. |
Which statement about the composition of vapour over an ideal `1 : 1` molar mixture of benzene and toluene is correct ? Assume the temperature is constant at `25^(@)C` Vapour pressure data `(25^(@)C)`: `{:("Benzene",75 mm Hg,),("Toluene",22 mm Hg,):}`A. The vapour will contain higher precentage of benzeneB. The vapour will contain higher precentage of tolueneC. The vapour will contain equal amount of benzene and tolueneD. Not enough information is given to make a prediction |
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Answer» Correct Answer - B `V = (W ("mass"))/(d) = (205.5)/(0.79) = 260.13 mL` `M = (w_(B) xx 1000)/(m_(B) xx V)` `= (5.5 xx 1000)/(36.5 xx 260.13) = 0.58 M` |
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| 1307. |
A flask is partially evecuated to 400 torr pressure of air. A small amount of benzene is introduced into the flask in order that some liwquid will remain after equilibrium has been established. The vapour pressure of at `25^(@)C` is 200 torr. What is the total pressure in the flask at equilibrium at `25^(@)C` ?A. 120 torrB. 510 torrC. 620 torrD. 480 torr |
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Answer» Correct Answer - c Since the temperature `(25^(@)C)` os constant the total vapour pressure of the equilibrum mixture= 400+220=620 torr. |
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| 1308. |
Which of the statements about the composition of the vapour over an ideal 1:1 molar mixture of benzene and toluene is correct ? Assume that the temperature is coinstant at `25^(@)C`. (Give, vapour pressure data at `25^(@)C`, benzene = 21.8 kPa, toluene = 3.85 kPa)A. The vapour phase will contain equal amounts of benzene and tollutene.B. Not enough information is given to make a predication.C. The vapour phase will contain a higher precentage of benzeene.D. The vapour phase will contain a higher percentage of toluene. |
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Answer» Correct Answer - c `P_("Benzene")=X_("Benzene")xxP_("Benzene")^(@)` =`1//2xx121.8 k Pa` `P_("tonene")=X_("toluene")xxP_("toluene")^(@)` =`1//2xx3.85=1.925kPa` Since `P_("Benzene")` is more than that of toluene the vapour phses will ciontain higher percentage of benzene |
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| 1309. |
The solubility of Ba(OH)2.H2O at 298 K is 5.6 g per 100 g of water. What is the molality of OH- ions in saturated solution? |
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Answer» m = Mass of solute (g) x 1000 /Molar mass of solute x Mass of Solvent(g) m (Ba(OH)2.H2O) = 5.6 x 1000 /189 x 94.4 = 0.314 m m ( OH-) = 2 x m (Ba(OH)2.H2O) = 0.314 x 2 = 0.628 m. |
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| 1310. |
Calculate the number of moles of methanol in `5` litres in its `2m` solution, if the density of the solution `0.981kgL^(-1)` (molar mass of methanol `=32g "mol"^(-1)`) |
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Answer» Correct Answer - `9.22` moles. `nXH_(3)OH=?` Mass of solution `=` Volume of solution `xx` density of solution Mass of solution `=5Lxx0.981kg l^(-1)=4.905kg=4905g` Mass of solute`=2xx32=64g`. Mass of `2m` solution containing `1kg` of solvent `=64+1000=1064g`. `4905g` of solution contains `=(2 "moles")/(164)xx4905=9.22` moles of methanol. |
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| 1311. |
Calculate the mass of Urea required in making 2.5 Kg of 0.25 molal aqueous solution. |
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Answer» moles of Urea=0.25 Molar mass of Urea=60 Mass of Urea=0.25 x 60=15 g Mass of solvent =1000 g Mass of solution=15+1000 = 1015 g 1015 g of solution has =15 g of Urea Hence 2500 g of aqueous solution has =15/1015 X 2500 g of urea =36.95 g. |
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| 1312. |
Calculate the number of moles of methanol in 5 litres of its 2 m solution if the density of solution is 0.981 kg (Molar mass of methanol=32.0g `mol^(-1)`. |
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Answer» 2 m solution of methanol means that 2 moples of methanol are dissoved in 1 kg (1000g) of water. Mass of methnol `(CH_(3)OH)` in 2 moles`=(2mol)xx(32.0gmol^(-1))=64.0kg` Dencity of solution =`0.981kgL^(-1)` Volume of solution `= (Mass)/(Density)=((1.064kg))/(0.981kgL^(-1))=1.085L` Now, 1.085 L of solution contain methanol =2mol 5.0L of soultion contain mehanol`=((2mol))/((1.085L))xx(5.0L)=9.22` mol. |
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| 1313. |
The solubility of `Ba(OH)_(2)`. `8H_(2)O` in water at `288K` is `5.6g` per `100g` of water. What is the molality hydroxide ions in saturated solution of `Ba(OH)_(2)`. `8H_(2)O` at `288K` ? [At. Mass of `Ba=137,O=16,H=1`) |
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Answer» Correct Answer - `0.356m` `2xx` moles of `Ba(OH)_(2)=` moles of `OH^(-)` `2xx(5.6)/(315)=` moles of `OH^(-)` molality of `OH=2xx(5.6)/(315)xx(1000)/(100)=0.356m`. |
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| 1314. |
The solubility of `ba(OH)_(2)8H_(2)O` in water at 298 K is 5.6 g per 100 g of water. What is the molality of the hydroxide ions in a saturatd solution of barium hydroxide at 298 K ? (Atomic mass of Ba =137, O=16) |
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Answer» `"Molality of solution(m)"=("Mass of solute"//"Molarmass of solute")/("Mass of solvent in kg")` Mass of solute, `Ba(HO)_(2)8H_(2)O=5.6g` Molar mass of solute, `Ba(OH)_(2)8H_(2)O=137+2(16+1)+8xx18=315 g mol^(-1)` Mass of solvent (water)= 100g=0.1 kg `Molality (m)=((5.6g)//(315gmol^(-1)))/((0.1kg))=0.18 molkg^(-1)=0.18m` Calculation of molality of the OH ions Barium hydroxide ionies inwater as follows: `Ba(OH)_(2) overset((aq)) Leftrightarrow Ba^(2)+(aq)+2OH(aq)` This shows that the molality of OH ions is twice the molality of `Ba(OH)_(2)=2xx0.18m=0.36m` |
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| 1315. |
What is the similarity between Raoult’s law and Henry’s law? |
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Answer» According to Raoult‘s law, the vapour pressure of a volatile componentin a given solution is given by pi = xi pi0. In the solution of a gas in aliquid, one of the components is so volatile that it exists as a gas and We have already seen that its solubility is given by Henry‘s law which states thatp = KHx. If we compare the equations for Raoult‘s law and Henry‘s law, itcan be seen that the partial pressure of the volatile component or gasis directly proportional to its mole fraction in solution. Only the proportionality constant KH differs from p10 . Thus, Raoult‘s law becomesa special case of Henry‘s law in which KH becomes equal to p10. |
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| 1316. |
State Raoult’s law for the solution containing two volatile liquids? |
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Answer» It states that the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. Thus for the component 1, P1= P1o X1 where P1 is the vapour pressure of the component 1 while P1 o is the vapour pressure of the same component in the pure state. The mole fraction of the component 1 is X1 |
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| 1317. |
Which salt shows maximum osmotic pressure in its `1 m` solution.A. `AgNO_(3)`B. `Na_(2)SO_(4)`C. `(NH_(4))(3)PO_(4)`D. `MgCl_(4)` |
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Answer» Correct Answer - C Osmotic pressure `prop` moles, `(NH_(4))_(3)PO_(4)` furnishes 4 ions in solution. |
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| 1318. |
What is meant by stabilizer; give example. |
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Answer» The soft drinks available in market do not settle down even after prolonged storage. To retain as such for long time without settling down some chemicals are added to it. These are called stabilizers. Eg. Brominates vegetable oil. |
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| 1319. |
The vapour pressure of a solution of a non-volatile electrolyte `B` in a solvent `A` is `95%` of the vapour pressure of the solvent at the same temperature. If the molecular weight of the solvent is `0.3` times, the molecular weight of solute, the weight ratio of the solvent and solute are:A. 0.15B. 5.7C. 0.2D. 4 |
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Answer» Correct Answer - B `"P"_("solution")=0.95"P"_("solvent")^(0)` ` "m"_("solvent")=0.3" m"_("solute")` `("P"_("solvent")^(0)-"P"_("solution"))/("P"_("solution"))=("W"_("solute")xx"m"_("solvent"))/("m"_("solute")xx"W"_("solvent"))` `("P"^(0)-0.95"P"^(0))/(0.95"P"^(0))="W"_("solute")/"W"_("solvent")xx0.3` `"W"_("solute")/"W"_("solvent")=0.3xx19=5.7` |
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| 1320. |
Two beakers of capacity 500 mL were taken. One of these beakers, labelled as "A" was filled with 400 mL water whereas the beaker labelled "B" was filled with 400 mL of 2M solution of NaCl. At the same temperature both the beakers ware placed in containers os same material and same capacity as shown in figure. A. Vapour pressure in container (A) is more than that of container (B)B. Vapour pressure in container (A) is less than that in container (B)C. Vapour pressur is equal in both the containersD. Vapour pressure in container (B) is twice the vapour pressure in container (A) |
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Answer» Correct Answer - A When salt is added to water to make the solution the vapour pressure of solution get decreasses. This is due to decrease in surface covered by solvent molecule which lead to decrease in number of solvent molecule escaping from the surface corresponding to pure solvent. Hence, vapour pressure also get reduces. |
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| 1321. |
What should be the boiling point of `1.0 molal` aqueous `KCl` solution (assuming complete dissociation of KCl) if `K_(b)^(H_(2)O)` is `0.52 K m^(-1)`?A. `100.52^(@)C`B. `101.04^(@)C`C. `99.48^(@)C`D. `98.96^(@)C` |
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Answer» Correct Answer - B `DeltaT_(b)=iK_(b)m = 2 xx 0.52 xx 1 = 1.04(@)C` rArr T_(b) = 100 + DeltaT_(b) =101.04^(@)C` |
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| 1322. |
`x` mole of `KCl` and `y` mole of `BaCl_(2)` are both dissolved in `1kg` of water. Given that `x+y=0.1` and `K_(f)` for water is `1.85K//"molal`, what is the observed range of `DeltaT_(f,)` if the ratio of `x` to `y` is varied ?A. `0.37^(0)` to `0.555^(0)`B. `0.185^(0)` to `0.93^(0)`C. `0.56^(0)` to `0.93^(0)`D. `0.37^(0)` to `093^(0)` |
| Answer» Correct Answer - A | |
| 1323. |
Calculate the molality of 10g of glucose in 90g of water. |
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Answer» `"Molality"=("Weight of solute"xx1000)/("GMW of solute "xx" weight of solvent in grams")` `=(10xx1000)/(180xx90)=0.6173" mol"//"Kg"` |
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| 1324. |
For a solution of `0.849g` of mercurous chloride in `50g` of `HgCl_(2)(l)` the freezing point depression is `1.24^(@)C. K_(f)` for `HgCl_(2)` is `34.3`. What is the state of mercurous chloride in `HgCl_(2)` ? `(Hg-200,CI-35.5)`A. as `Hg_(2)Cl_(2)` moleculesB. as `HgCl` moleculesC. as `Hg^(+)` and `Cl^(-)` ionsD. as `Hg_(2)^(2+)` and`Cl^(-)` ions |
| Answer» Correct Answer - A | |
| 1325. |
The hardness of water sample containing `0*002` mole of magnesium sulphate dissolved in a litre of water is expressed asA. 20 ppmB. 200 ppmC. 2000 ppmD. 120 ppm |
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Answer» Correct Answer - B `10^(6)` g water contains `(xx(0.002xx1000)"mol MgSO"_(4)` 1 mol `MgSO_(4)~=1` mol `CaCO_(3)` `:.` 2 mol `MgSO_(4)~= 2` mol `CaCO_(3)`, i.e., `2 xx 100 g CaCO_(3)` `:.` Hardness of water = 200 ppm] |
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| 1326. |
Consider separate solution of `0.500M C_(2)H_(5)OH(aq), 0.100M Mg_(3)(PO_(4))_(2)(aq), 0.250M KBr(aq)` and `0.125 M Na_(3)PO_(4)(aq)` at `25^(@)C`. Which statement is true about these solutions, assuming all salts to be strong electrolytes?A. `0.125 M Na_(3)PO_(4)(aq)` has the highest osmotic pressureB. `0.500 M C_(2)H_(5)OH(aq)` has the highest osmotic pressureC. They all have the same osmotic pressureD. `0.100M Mg_(3)(PO_(4))_(2)(aq)` has the highest osmotic pressure |
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Answer» Correct Answer - C `pi =iCRT` `|{:(S.No .,"Electrolyte",i,C,ixxC),(1,C_(2)H_(6)OH,1,0.5M,0.5),(2,Mg_(3)(PO_(4))_(2),5,0.1M,0.5),(3,KBr,2,0.25M,0.5),(4,Na_(3)PO_(4),4,0.125M,0.5):}:|` `:.` all have same osmotic pressure |
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| 1327. |
What is the mount of KCI in grams which must be added in 100 g of water so that the freezing point is depressed by 2 K ? `(K_(f)"for water "= 1.86 K/m)`. |
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Answer» `W_(B)=(DeltaT_(f)xxM_(B)xxW_(A))/(iK_(f))` `=((2K)xx(74.5"g mol"^(-1))xx(0.1kg))/(2xx(1.86"K kg mol"^(-1)))=4` |
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| 1328. |
How maby grams of methyl alcohol should be added to 10 litre tank of water to prevent its freezing at `268K`? `(K_(f)` for water is `1.86 K kg mol^(-1))`A. `899.04g`B. `886.02g`C. `868.06g`D. `880.07g` |
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Answer» Correct Answer - C `DeltaT_(f) = K_(f) xx m` `5 = 1.86 xx (W_(A))/(32 xx 10)` `W = (5 xx 32 xx 10)/(1.86)` |
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| 1329. |
At` 25^@ `, the density of ` 15 M H_(2)SO_(4) `is ` 1.8 g cm^(-3) ` .Thus mass percentage of ` H_(2)SO_(4) ` in aqueous solution isA. ` 2% `B. ` 81.6% `C. ` 18% `D. ` 1.8% ` |
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Answer» Correct Answer - B `18 M H_(2)SO_(4)`means `18 mol L^(-1) 1000 mL of H_(2)SO_(4)` solution has`H_(2)SO_(4) = 18 mol =15xx98 g` Also, `1000 mL = 1000 xx 1.8 g H_(2)SO_(4)` solution thus ,`(1000xx1.8)g H_(2)SO_(4)` solution has `H_(2)SO_(4) = 15xx98 g` `100 g H_(2)SO_(4)` solution has `H_(2)SO_(4) = (15xx98)/(1000xx1.8)=(1470)/(1800)=81.6%` Thus , mass percentage of` H_(2)SO_(4) = 81.6%` |
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| 1330. |
3. 5 × 10-3 kg of urea is dissolved in 2 × 10-2 kg of water. The percentage by weight of urea is :(a) 15% (b) 20% (c) 25% (d) 30% |
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Answer» Option : (b) 20% |
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| 1331. |
`5 ml` of `NHCL ,20 ml` of `N//2` of `H_(2)SO_(4)` and `30 ml` of `N//3 HNO_(3)` are mixed together and volume made to one litre.A. ` (N)/(40) `B. ` (N)/(10) `C. ` (N)/(20) `D. ` (N)/(5) ` |
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Answer» Correct Answer - A `NV =N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3) ` or , ` 1000 = 1xx+(1)/(2)xx20+(1)/(3)30 or N =(1)/(40). ` |
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| 1332. |
Hardness of a water sample is ` 100 "ppm" CaCo_(3) `. Thus molarity of ` CaCO_(3) `isA. ` 2xx10^(-3)M `B. ` 1xx10^(-3) M `C. ` 2xx10^(-2)M `D. ` 2xx10^(-4) M` |
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Answer» Correct Answer - B Hard water has hardness `= 100 "ppm"` Thus , `10^(6) ml (=10^(3))` water has `CaCO_(3) = 100 g = 1 mol` Thus, molarity `= ("Moles")/("Litre") = (1)/(10^(3))= 1xx10^(-3) M` |
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| 1333. |
Calculate the amount of `KCl` which must be added to `1kg` of water so that the freezing point is depressed by `2K`. (`K_(f)` for water `=1.86K kg "mol"^(-1)`). |
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Answer» Correct Answer - `KCIrarrK^(+)+Cl^(-)rArri=2` `DeltaT_(f)=iK_(f)xxm` `2K=2xx1.86xxm rArrm=(1)/(1.86)` `m=0.5376 "mol"//kg` Amount of `KCl=0.5376xx74.5=40.05g`. |
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| 1334. |
`0.1 M` aqueous solution of `MgCl_(2)` at `300 K` is `4.92 atm`. What will be the percentage ionination of the salt? a.`49%` , b. `59%` ,c.`79%` d. `69%` |
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Answer» a.`pi=iCRT` `4.92=ixx0.1xx0.0821xx300` `i=1.99` `alpha=(i-1)/(n-1)=(1.99-1)/(3-1)=(0.99)/(2)=0.49` Perecentage of ionization=`49%` |
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| 1335. |
The molality of a sulphuric acid in which the mole fraction of water is 0.86 is......... |
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Answer» Correct Answer - `rarr 6` Mole fraction of `H_(2)SO_(4)` in solution `=1 - 0.86 = 0.14` `= (n_(2))/(n_(1)+n_(2)) = 0.14` Molality is `n_(2)` is 1000g of water i.e., `n_(1)` `= 1000//18 = 55.55` moles. `(n_(2))/(55.55+n_(2)) = 0.14` or `n_(2) = 0.14 n_(2) +7.77` or `0.86 n_(2) = 7.77` or `n_(2) = 9`. |
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| 1336. |
Which is better method for expressing concentration of solution – Molarity or Molality? |
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Answer» Molality = moles of solute/ volume of solution in L Molality = Moles of solute/Mass of the solvent in kg Volume changes with temperature whereas mass does not change with temperature. So molality, which does not have volume term in it is better method for expressing concentration. |
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| 1337. |
The temperature at which molarity of pure water is equal to its molalityA. `273K`B. `298K`C. `277K`D. none of these |
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Answer» Correct Answer - C At `4^(@)C (277K)` density of water is `1g cm^(-3)` `:. 1000 cm^(3)` of water `= 1000g` of water. `:.` At 277 K molality of pure water is same as its molarity (55.56). |
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| 1338. |
Assertion : Molality is perferred to molarity for expression the concentration of the solution. Reason : Molality can be determined more easily than molarity.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 - c Correct Reason. Molality does not change with temperature while molarity changes. |
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| 1339. |
The osmotic pressure of the blood cells is approximately equal to at 37°C. |
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Answer» reverse osmosis is applied in water purification. |
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| 1340. |
What happens when blood cells are placed in pure water? |
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Answer» Blood cells swell up due to osmosis (plasmolysis). |
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| 1341. |
From a measurement of the freezing point depression of benzene, the molecular weight of acetic acid in a benzene solution was determined to be `100`. The percentage association of acetic acid isA. `100%`B. `79%`C. `93%`D. `83%` |
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Answer» Correct Answer - 2 We can find degree of association as follows: `alpha=(M_(t)-M_(o))/M_(o)(n/(1-n))` where `M_(o)` is the observed molar mass and `M_(t)` is the theoretical molar mass. Here, we have `M_(o)=100 g//"mole"` `M_(t)=(12xx2+1xx4+16xx2)=60 g mol^(-1)` `n=2` Substituting these values, we get `alpha=(60-100)/100(2/(1-2))` `=80/100=0.8` percentage association `=0.8xx100%` `=80%` |
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| 1342. |
A 1.00 m solution of acetic acid (CH3COOH) in benzene has a freezing point depression of 2.6 K. Calculate the value for i and suggest an explanation for its value. |
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Answer» ΔTf = i x Kf x m 2.6 = i x 5.12 x 1 i = 2.6 / 5.12 = 0.502 I < 1 the result suggest that acetic acid undergoes association in organic solvent . |
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| 1343. |
From a measurement of the freezing point depression of benzene, the molecular weight of acetic acid in a benzene solution was determined to be `100`. The percentage association of acetic acid isA. `79%`B. `93%`C. `80%`D. `100%` |
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Answer» Correct Answer - C `i=60/100=1-alpha/2 rArr alpha= 80%` |
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| 1344. |
What is the optimum concentration of fluoride ions for cleaning of tooth? |
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Answer» The optimum concentration of fluoride ions for the cleaning of tooth is 1.5 ppm. [If it is more than 1.5 ppm it can be poisonous and if less than 1.5 ppm it is in effective.] |
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| 1345. |
What weight of glycerol should be added to 600 g of water in order to lower its freezing point by `10^@ C` ? `(K_f=1.86^@ C m^(-1))`A. 496 gB. 297 gC. 310 gD. 426 g |
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Answer» Correct Answer - B `DeltaT_f=K_fxxm=(K_fxxW_2xx1000)/(M_2xxW_1)` `W_2=(10xx92xx600)/(1.86xx1000)`=296.77 g |
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| 1346. |
What will be the freezing point of a 0.5 m KCl solution ? The molal freezing point constant of water is `1.86^@C m^(-1)` .A. `-1.86^@C`B. `-0.372^@C`C. `-3.2^@C`D. `0^@C` |
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Answer» Correct Answer - A `DeltaT_f=iK_fxxm=2xx1.86xx0.5=1.86^@C` `T_f=T_f^@-DeltaT_f=0-1.86=-1.86^@C` |
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| 1347. |
Define cryoscopic constant (or molal depression constant). |
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Answer» Molal depression constant : It is defined as the depression in freezing point, produced by dissolving one mole of a solute in 1 kg (or 1000 g) of a solvent (i.e. 1 molal solution). |
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| 1348. |
A saturated solution of `Mg(OH)_(2)` has a vapour pressure of `759.5 mm` at `373 K`. Calculate the solubility and `K_(sq)` of `Mg(OH)_(2)`. "(Assume molarity equals molality)" |
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Answer» `P_(S)=759.5 mm` `P^(@)_(H_(2)O)=760 mm` at `373 K` `DeltaP/P^(@)=0.5/760 =ichi_(2)` `:.ichi(2)=6.5xx10^(-4)(i=3)` `chi_2=2.16xx10^-4` Find molality`~~`molarity `m=(chi_(B)xx1000)/(1-chi_(B)xxMw_(1))=0.012 mol kg^(-1)` `Mg(OH)_2 hArr Mg^(2+)+2OH^(ө)` `i=3` `S=0.012 mol L^(-1)` `K_(sp)=4S^(3)` `=4xx(0.0112)^(3)` `=6.8xx10^(-6)` |
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| 1349. |
Write the formula to determine molar mass of a solute using freezing point depression method. |
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Answer» M2 = \(\frac{K_f\times W_2 \times 1000}{W_1\times ΔT_f}\) where Kf = Molal depression constant ΔTf = Depression in freezing point W1 = Mass of a solvent W2 = Mass of a solute. M2 = Molar mass of solute |
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| 1350. |
A mixture of `0.1 mol ` of `Na_(2)O` and `0.1 mol ` of `BaO` is dissolved in `1000 g` of `H_(2)O`. Calculate the vapour pressure of solution at `373 K`. |
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Answer» `Na_(2)O` and `BaO` both are ionic compounds and are completely ionized to give `3` and `2 ions`, respectively, `(i) Na_(2)O rarr 2Na^(o+) O^(2)-(i=3)` `(ii)BaO rarr Ba^(2+) + O^(2-) (i=2)` `:.(P^(@)-P_(s))/P^(@) = ichi_(Na_(2)O) + ichi_(BaO)` `(760-P_(s))/(760) =3xx(0.1)/(0.1+0.1+55.5)+2xx (0.1)/(0.1+0.1+55.5)` `:.P_(s)=753.21 mm Hg` Alternatively Total number of ions=`5xx0.1=0.5` `(P^(@)-P_(s))/P^(s) =n_(2)/n_(1) =0.5/55.5 =0.009` `760-P_(s)/P_(s)=0.009` Solve for `P_(s) rArr 753.21 mm Hg`. |
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