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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.
| 1351. |
Which of the following exhibits lowest boiling point?A. `0.1M NaCI`B. `0.1M CaCI_(2)`C. `0.1M CH_(3)COONa`D. `0.1M` glucose |
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Answer» Correct Answer - D More the number of solute particles more is the elevation in boiling point. Lowest boiling point implies least elevation in boiling point. This is possible when number of solute particles is least i.e., in ease of glucose. |
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| 1352. |
How is molar mass related to the elevation in boiling point of a solution ? |
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Answer» The expression for elevation of boiling point is `DeltaT_(b)=(K_(b)xx1000xx w)/(m xx w)=K_(b)=" molal elevation constant"` w = Weight of solute W = Weight of solvent m = molar mass of solute Molar mass of solute m `=(K_(b)xx1000xx w)/(DeltaT_(b)xx w)` `:.` Molar mass of solute (m) and elevation of boiling point `(DeltaT_(b))` are inversely related. |
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| 1353. |
Calculate the mass of urea `("NH"_(2)"CONH"_(2))` required in making 2.5 kg of 0.25 molar aqueous solution. We know that molarity (m) `=("Moles of solute")/("Mass of solvent in kg")` and moles of soute `=("Mass of solute")/("Molar mass of solute")` So, find the molar mass of solute by adding atomic masses of different element present in it and mass by using the formula, Molality `=("Mass of solute/molar mass of solute")/("Mass of solvent in kg")` |
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Answer» Molality of the solution `=0.25" m"=0.25" mol kg"^(-1)` Molar mass of urea `("NH"_(2)"CONH"_(2))=(14xx2)+(1xx4)+12+16` `=60" g mol"^(-1)` Mass of solvent (water) = 2.5 kg Molality `=("Mass of urea/molar mass of urea")/("Mass of water in kg")` `(0.25" mol kg"^(-1))=("Mass of urea")/((60" g mol"^(-1))xx(2.5" kg"))` Mass of urea `=(0.25" mol"^(-1))xx(60" g mol"^(-1))xx(2.5" kg")=37.5" g"` |
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| 1354. |
You are supplied with supplied with 500mL each of 2NHCL and HCL solutions.What is the maximum volume of 2M HCL solution that you can prepare using only these two solution ? |
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Answer» For HCI, Molarity=Normality. Let x mL be the maximum volume of 3M HCL that can be prepared by mixing 500 mL of 2MHCL solution and same volume of 5M HCI solution. `M_(1)V_(1)+M_(2)V_(2)=M_(3)V_(3)` `(2Mxx500mL)+(5MxxxmL)=3Mxx(500+x)mL` (100+5x)=(1500+3x) 2x= 1500-1000=500orx=250mL Thus , the maximum volume of 3M solution that can be prepared =(500+250)=750 mL |
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| 1355. |
Calculate the molarity of each of the following solution : (a) 30g of `CO(NH_(3))_(2).6H_(2)O` in 4.3 L of solution. (b) 30 mL of `0.5" MH"_(2)"SO"_(4)` diluted to 500 mL. (a) Molarity `=("moles of solute")/("Volume of solution litre")` and moles of solute `=("mass of solute")/("molar solution of solute")` So, first find molar mass by adding atomic masses of different elements, then find moles of solute and then molarity. (b) Use molarity equation for dilution. `M_(1)V_(1)=M_(2)V_(2)` (Before dilution) (After dilution) |
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Answer» Molar mass of `CO(NO_(3))_(2).6H_(2)O=(58.7)+2(14+48)+(6xx18)"g mol"^(-1)` `=58.7+124+108=290.7~~291" g mol"^(-1)` Mole of `CO(NO_(3))_(2)6" H"_(2)O=(30"g")/(291" g mol"^(-1))=0.103` Volume of solution = 4.3 L Molarity (M) `=(0.103" mol")/(4.3" L")=0.024" mol L"^(-1)=0.024" M"` (b) Volume of undiluted `H_(2)SO_(4)` solution `(V_(1))=30" mL"` Molarity of undiluted `H_(2)SO_(4)` solution `(M_(1))=0.5" M"` Volume of diluted `H_(2)SO_(4)` solution `(V_(2))=500" mL"` We know that `M_(1)V_(1)=M_(2)V_(2)` `:. M_(2)=(M_(1)V_(1))/(V_(2))=((0.5"M")(30" mL"))/(500" mL")=0.03" M"` |
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| 1356. |
Calculate the molarity of each of the following solutions: 30g of `Co(NO_(3))_(2)6H_(2)O` in 4.3Lpf solution `30 mL of 0.5 M H_(2)SO_(4)`diluted to 500 mL |
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Answer» Molarity of soluion`=("Mass of solute"//"Molar mass of solute")/("Volume of solution in litres")` Mass ofsolute, `Co(NO_(3))_(2).6H_(2)O=30g` Molar mass of solute, `Co(NO_(3))_(2)6H_(2)O=59+2xx14+6xx16+6xx18=291gmol^(-1)` Volume of solution=4.3L Molarity (M)`=((30g)//(291gmol^(-1)))/((4.3L))=0.024 M` Volume of undiluted `H_(2)SO_(4) "solution" (V_(1))=30mL` Molarity of undiluted `H_(2)SO_(4) "solution" (M_(1))=0.5M` Volume of diluted `H_(2)SO_(4) "solution" (V_(2))=500mL` Molarity of diluted `H_(2)SO_(4)(M_(2))` can be calculated as : `M_(1)V_(2)=M_(2)V_(2)` `M_(2)=(M_(1)V_(1))/V_(2)=((30mL)xx(0.5M))/((500mL))=0.03M` |
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| 1357. |
The vapour pressure of pure benzene, `C_(6)H_(6)` at `50^(@)C` is 268` Torr. How many moles of non-volatile solute per mol of benzene is required to prepare a solution of benzene is required to prepare a solution of benzenen having a vapour pressure of `167` Torr at `50^(@)C` ?A. `0.377`B. `0.605`C. `0.623`D. `0.395` |
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Answer» Correct Answer - A `(268-167)/(268)=xrArr So xx=0.37`. |
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| 1358. |
`V.P.` of solute containing `6gm` of non volatile solute in `180gm` of water is `20 "Torr"//mm` of `Hg`. If `1` mole water is further added in to the `V.P.` increase by `0.02`. Torr calculate `V.P.` of pure water `&` molecular of non volatile solute. |
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Answer» `(P^(0)-P_(s))/(P_(s))=(w)/(m)xx(M)/(W)` `rArr (P^(0)-20)/(20)=(6)/(m)xx(18)/(180)` `(P^(@)-20.02)/(20.02)=(6)/(m)xx(18)/(198)` `rArr P^(@)=20.22` Torr. `m=54 gm//mol`. |
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| 1359. |
Define mass percentage, volume percentage and mass to volume percentage solutions. |
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Answer» i) Mass percentage : The mass percentage of a component of a solution is defined as the `"mass"% =("Mass of the component in the solution")/("Total mass of the solution")xx100` ii) Volume percentage `((v)/(V))` : It is defined as the volume % of a component `=("Volume of the component")/("Total volume of solution")xx100` iii) Mass to volume percentage `((w)/(V))` : It is defined as the mass of solute dissolved in 100 ml of the solution. `to` It is commonly used in medicine and pharmacy. |
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| 1360. |
Calculate the weight of Glucose required to prepare 500ml of 0.1 M solution. |
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Answer» Molarity = 0.1, Molecular weight of glucose = 180, Volume of solution = 500 ml Weight of solute `=("Molarity "xx" Gram molecular weight of solute "xx" vol of solin ml")/(1000)` Weight of solute `=(0.1xx180xx500)/(1000)=9" g ":." Weight of glucose"=9" g"` |
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| 1361. |
How many litres of`CO_(2)`atSTP will be formed when `100ml`of `0.1MH_(2)SO_(4)`reacts with excess of `Na_(2)SO_(3)` ?A. `22.4`B. `2.24`C. `0.224`D. `5.6` |
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Answer» Correct Answer - C `Na_(2)CO_(3)+underset(98 "gram(2 mole)")(H_(2)SO_(4))rarr underset("1 mole")(Na_(2)SO_(4))+underset("1 mole")(CO_(2)Ho)` `0.02=(0.02xx22.4)/(2)=0.224` |
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| 1362. |
A solution of glycerol `(C_(3)H_(8)O_(3))` with mlar mass =92 g `mol^(-1)` in water was prepared by dissolved some glycerol in 500 g of water. The solution had a boiling point of `100.42^(@)C`. What mas of glycerol was dissolved to prepare this olution (`K_(2)` for water = 0.512 K kg `mol^(-1)`). |
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Answer» Correct Answer - 37.7 g `DeltaT_(b)=(100.42^(@)C-100)^(@)C=0.42^(@)C=0.42" K",` `M_(B)=92" g mol"^(-1),W_(A)=500 g =0.5" kg",` `K_(b)=0.152" k kg mol"^(-1)` `W_(B)=(M_(B)xxDeltaT_(b)xxW_(A))/(K_(b))=((92g mol^(-1))xx(0.42K)xx(0.5kg))/((0.512" k kg mol"^(-1)))=37.7 g`. |
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| 1363. |
The boliling point of 5% solution (W/W) of non-volatile solute in water is `100.45^(@)C`. The boiling point of pure water is `100^(@)C`.Calculate th molar mas of th solute `(K_(b)"for water=0.52 K kg mol"^(-1))`. |
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Answer» Correct Answer - 60.8 g `mol^(-1)` `DeltaT_(b)=100.45^(@)C-100^(@)C=0.45 K` `W_(B)=5g, W_(A)=100-5=95g, 0.095 kg, K_(b)=0.52" K kg mol"^(-1)` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.52K" kg mol"^(-1))xx(5g))/((0.45K)xx(0.095 kg))=60.8" g mol"^(-1)`. |
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| 1364. |
The boiling point of water is `100^(@)C` and it becomes `100.52^(@)C` if 3 g of a non-voltile is dissolved in 20 mL of it. Calculate the molecular weight of the solute. (`K_(b)` for water is 0.52 K kg `mol^(-1)`, density of water= 1 g `mol^(-1)`). |
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Answer» Correct Answer - 150 g `mol^(-1)` `DeltaT_(b)=100.52-100=0.52^(@)C=0.52 K` `W_(B)= 3 g, W_(A)=20 g [20" mL of water= 20 g"], 0.02 kg, k_(b)=0.52" K kg mol"^(-1)` `M_(B)= ?` `M_(B)=(K_(b)xxW_(b))/(DeltaT_(b)xxW_(A))=((0.52" K kg mol"^(-1))xx(3g))/((0.52K)xx(0.02 kg))=150" g mol"^(-1)`. |
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| 1365. |
The boiling point of water becomes `100.52^(@)C` if 1.5 g of a non-volalite solute is dissolved in 100 mL of it. Calculate the molecular weight of the solute. (`K_(b)` for water=0.6 K kg `mol^(-1)`). (density of waer-1 g `moL^(-1)`). |
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Answer» Correct Answer - 17.31 g `mol^(-1)` `DeltaT_(b)=100.52-100=0.52^(@)C=0.52 K` `W_(B)=1.5g, W_(A)=100g=0.1 kg " "[1" mL of water"~~1g]` `K_(b)=0.6K//m, M_(B)=?` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.6" K kg mol"^(-1))xx(1.5g))/((0.52 K)xx(0.1 kg))=17.31" g mol"^(-1)` |
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| 1366. |
Calculate the molecular mass of a substance, 1.0 gram of which when dissolved in 100 gram of solvent gace an elecation of 0.307 K in the boiling point. (Molar elevation constant `K_(b)=1.84 K kg mol^(-1)`). |
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Answer» Correct Answer - 59.93 g `mol^(-1)` `W_(B)=1.0g, W_(A)=0.1 kg, DeltaT_(b)=0.307 K, K_(b)=1.84 K//m, M_(B)=?` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((1.84" K kg mol"^(-1))(1.0g))/((0.307 K)xx(0.1 kg))=59.93" g mol"^(-1)`. |
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| 1367. |
A certain vessel `X` has water and nitrogen gas at a total pressure of 2 atm and `300 K`. All the contents of vessel and transferred to another vessel `Y` having half the capacity of the vessel `X`. The pressure of `N_(2)` in this vessel was 3.8 atm at `300 K`. The vessel `Y` is heated to `320 K` and the total pressure observed was 4.32 atm. Assume that the volume occupied by the gases in vessel is equal to the volume of the vessel. Calculate the following: Pressure of water vapour at `320 K`.A. `0.27`B. `0.32`C. `4.0`D. `1.0` |
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Answer» Correct Answer - C Total Pressure at `320 K =4.32` atm `:.` Pressure of water vapour at `320 K =4.32-4.05` =0.27 atm |
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| 1368. |
A certain vessel `X` has water and nitrogen gas at a total pressure of 2 atm and `300 K`. All the contents of vessel and transferred to another vessel `Y` having half the capacity of the vessel `X`. The pressure of `N_(2)` in this vessel was 3.8 atm at `300 K`. The vessel `Y` is heated to `320 K` and the total pressure observed was 4.32 atm. Assume that the volume occupied by the gases in vessel is equal to the volume of the vessel. Calculate the following: Pressure of `H_(2)O` in `X` at `320 K`.A. `0.1`B. `0.2`C. `1.0`D. `2.0` |
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Answer» Correct Answer - A Pressure of `N_(2)` in `X = 1.9 atm Total pressure of vessel X=2 atm `:.` Pressure of `H_(2)O(g)` in `X` at `300 K =2-1.9 =0.1 "atm"` |
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| 1369. |
A certain vessel `X` has water and nitrogen gas at a total pressure of 2 atm and `300 K`. All the contents of vessel and transferred to another vessel `Y` having half the capacity of the vessel `X`. The pressure of `N_(2)` in this vessel was 3.8 atm at `300 K`. The vessel `Y` is heated to `320 K` and the total pressure observed was 4.32 atm. Assume that the volume occupied by the gases in vessel is equal to the volume of the vessel. Calculate the following: Pressure of `H_(2)` at `320 K`.A. `4.0`B. `4.05`C. `5.05`D. `1.05` |
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Answer» Correct Answer - B Pressure of `N_(2)` in Y = 3.8 atm Pressure of `N_(2)` at `320 K =3.8/300 x 320 =4.05` atm |
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| 1370. |
A certain vessel `X` has water and nitrogen gas at a total pressure of 2 atm and `300 K`. All the contents of vessel and transferred to another vessel `Y` having half the capacity of the vessel `X`. The pressure of `N_(2)` in this vessel was 3.8 atm at `300 K`. The vessel `Y` is heated to `320 K` and the total pressure observed was 4.32 atm. Assume that the volume occupied by the gases in vessel is equal to the volume of the vessel. Calculate the following: Enthalpy of vapourization.A. `30.00`B. `35.65`C. `38.65`D. `39.65` |
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Answer» Correct Answer - D Enthalpy of vapourization `(Delta_(vap)H)` `rArr log p_(1)/p_(2)=(DeltaH)/(2.303 R)[1/300 - 1/320]` `rArr log 0.27/0.10=(DeltaH)/(2.303 R)[1/300-1/320]` = `Delta_(vap)H=39.65 KJ "mol"^(-1)` |
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| 1371. |
A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendercy to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point `(DeltaT_(b))` and depression in freezing point `(DeltaT_(f))` of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, `(DeltaT_(b))` and `(DeltaT_(f))` are proportional to the molarity of the solute in the solution. Dissolution of a non-volatile solute into a liquid leads toA. A decrease of entropyB. An increase in tendency of the liquid to freezeC. An increases in tendency to pass into the vapour phaseD. A decrease in tendency of the liquid to freeze |
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Answer» Correct Answer - D Addition of non-volatile solutes decreases the freezing point of solution. |
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| 1372. |
A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendercy to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point `(DeltaT_(b))` and depression in freezing point `(DeltaT_(f))` of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, `(DeltaT_(b))` and `(DeltaT_(f))` are proportional to the molarity of the solute in the solution. A mixture of two immiscible liquids at a constant pressure of `1.0 atm` boils at temperatureA. Equal to the normal boiling point of more volatile liquid.B. Equal to the mean of the normal boiling points of the two liquids.C. Greater than normal boiling point of either of liquids.D. Smaller than the normal boiling point of either of the liquid. |
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Answer» Correct Answer - D Mixture will boil at lower temperature of the boiling point of either liquid. |
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| 1373. |
A one litre solution is prepared by dissolving some lead-nitrate in water. The solution was found to boil at `100.15^(@)C`. To the resulting solution `0.2` mole NaCI was added. The resulting solution was found to freeze at `0.83^(@)C`. Determine solubility product of `PbCI_(2)`. Given `K_(b) = 0.5` and `K_(f) = 1.86`. Assume molality to be equal to molarity in all case. |
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Answer» Correct Answer - `1.46 xx 10^(-5)` `Delta T_(b) = iK_(b)m` `0.15=3xx0.5xxm implies m=0.1` Now, `{:(Pb(NO_(3))_(2),+,2NaCl,rarr,PbCl_(2),+,2NaNO_(3)),(0.1,,0.2,,,,),(,,,,0.1,,0.2):}` Now, this solution constains two salts `Delta T_(f)=K_(f)xx m 0.83=1.86[2xx0.2+3s]` where s is molar solubility of `PbCl_(2)`. `s = 1.54xx10^(-3) K_(sp) = 4s^(3)=1.46xx10^(-5)` |
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| 1374. |
At `10^(@)C`, the osmotic pressure of urea solution is `500 mm`.The solution is diluted and the temperature is raised to `25^(@)C`.when the osmotic pressure is found to be `105.3 mm`. Determine the extent of dilution. |
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Answer» For initial solution,`pi=500/760`atm,`T=283 K` Using `piV=nRT rArr 500/760xx(V_(1))=nxxRxx283` …(i) After dilution, let volume becomes `V_(2)` and temperature is raised to `25^(@)C,i.e., 298 K`. `pi=105.3/760`atm `105.3/760xx(V_(2))=nxxRxx298` ...(ii) `:.` By Eqs.(i) and Eqs.(ii), we get `V_(1)/V_(2)=(283)/(298)xx(105.3)/(500)` `V_(1)/V_(2)=1/5 rArr V_(2)=5V_(1)` i.e., the solution was diluted to 5 times. |
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| 1375. |
Camphor is often used in molecular mass determination becauseA. it is reading availableB. it has a very high cryoscopic constantC. it is volatileD. it is a solvent for organic substances |
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Answer» Correct Answer - 2 We have `DeltaT_(f)=iK_(f)m` For a given concentration of a solution, the magnitude of `DeltaT_(f)` is directly related to the value of `K_(f)`. Among the available solvents, camphor has the highest value of `K_(f)(40 K kg mol^(-1))` and thus gives the maximum depression of freezing point for a given amount of solute. |
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| 1376. |
Ethylene glycol is added to water as an antifreeze. It willA. Decrease the freezing point of water in the winter and increase the boiling poing of water in the summerB. Only decrease the freezing point of waterC. Only increase the boiling point of waterD. Be used for cleaning the radiator in a car |
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Answer» Correct Answer - A Ethylene glycol is a non-volatile solute. It increases the b.pt. of water (thus, radiator coolant needs less topping up) and decreases the f.pt. (no freezing of radiator coolant in winter). |
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| 1377. |
The osomotic pressure `pi` depends on the molar concentration of the solution `(pi=CRT)`. If two solutions are of equal solute concentration and, hence, have the same omotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more diluted solution is described as hypotonic. Osmosis is the major mechanism for transporting water upward in the plants. Answer the following questions: A plant cell shrinks when it is kept in:A. Hypotonic solutionB. Hypertonic solutionC. Isotonic solutionD. Pure water |
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Answer» Correct Answer - C Hypertonic solutions have higher osmotic pressure and therefore they have higher concentration. When a plant cell is kept in hypertonic solution, water from plant cell moves to hypertonic solution and therefore the plant cell gets shrinked. |
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| 1378. |
The osomotic pressure `pi` depends on the molar concentration of the solution `(pi=CRT)`. If two solutions are of equal solute concentration and, hence, have the same omotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more diluted solution is described as hypotonic. Osmosis is the major mechanism for transporting water upward in the plants. Answer the following questions: What would be the percent strength of solution of urea that would be isotonic with `4.5%` solution of glucose?A. `4.5%`B. `13.5%`C. `1.5%`D. `9%` |
| Answer» Correct Answer - C | |
| 1379. |
Camphor is often used in molecular mass determination becauseA. it is solvent for organic substancesB. it is readily availableC. it has a very high cryoscopic constantD. it is volatile |
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Answer» Correct Answer - C Camphor has very large `K_(f)` (cryoscopic constant). As such, `Delta T_(f)` for camphor is very large and easyto find accurately. |
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| 1380. |
The osomotic pressure `pi` depends on the molar concentration of the solution `(pi=CRT)`. If two solutions are of equal solute concentration and, hence, have the same omotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more diluted solution is described as hypotonic. Osmosis is the major mechanism for transporting water upward in the plants. Answer the following questions: The glucose solution to be injected into the bloodstream and the blood itself should have the same.A. "Molarity"B. "Vapour pressure"C. "Osmotic pressure"D. "Viscosity" |
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Answer» Correct Answer - B The osmotic pressure should be same for glucose solution and bloodstream otherwise veins would get shrinked. |
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| 1381. |
The average osomotic pressure of human blood is 7.8 bar at `37^(@)C`. What is the concentration of an aqueous `NaCI` solution that could be used in the blood stream?A. `0.16 mol//L`B. `0.32 mol//L`C. `0.60mol//L`D. `0.45 mol//L` |
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Answer» Correct Answer - A `pi=7.8" bar ", T = 37^(@)C = 300 K` `R = 0.083L " bar " mol^(-1)K^(-1)` i = 2 for `Na^(+) Cl^(-)` `pi = (I xx n_(B)xx R xx T)/(V)` `(n_(B))/(V)= (pi)/(iRT)` `=(7.8"bar")/(2xx(0.083 L^(-1)mol^(-1)K^(-1))xx300K)` `= 0.1566 mol L^(-) = 0.16 mol/L` |
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| 1382. |
The empirical formula of a non-electrolyte is `CH_(2)O`. A solution containing 3 g `L^(-1)` of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution. The molecular formula of the compound is :A. `CH_(2)O`B. `C_(2)H_(4)O_(2)`C. `C_(4)H_(8)O_(4)`D. `C_(3)H_(6)O_(3)` |
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Answer» Correct Answer - B Concentration of compouinds in solution `= 3g L^(-) = (3)/(M) mol L^(-1)` Since it is isotonic with `0.015M` glucose solution `(3)/(M) = 0.05` or `M = 60` Empirical formula mass of `CH_(2)O = 30` `n = ("Mol .mass")/(EF"mass") = (60)/(30) = 2` `:.` molecular formula `= 2 xx CH_(2)O = C_(2)H_(4)O_(2)` |
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| 1383. |
200 mL of an aqueous solution of a protein contains its 1.26g. The osmotic pressure of this solution at 300K is found to be `2.57 xx 10^(-3)` bar. The molar mass of protein will be `(R = 0.083 L bar mol^(-1)K^(-1))`A. `51022 g mol^(-1)`B. `122044 g mol^(-1)`C. `31011 g mol^(-1)`D. `61038 g mol^(-1)` |
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Answer» Correct Answer - D `pi = cRT = [(W_(2))/(M_(2)) xx (1)/(V)] RT` or `M_(2) = (W_(2)RT)/(piV)` `= (1.26 g xx 0.083 L "bar" mol^(-1)K^(-1) xx 300 K)/(2.57 xx 10^(3)"bar" xx 0.200L)` `= 61038 g mol^(-1)`. |
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| 1384. |
`200cm^(3)` of an aqueous solution contains 1.26g of a polymer. The osmotic pressure of such solution at 300K is found to be `2.57 xx 10^(-3)` bar. Calculate the molar mass of the polymer. |
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Answer» `piV=(w_(B))/(m_(B))RT` `m_(B)=(w_(B))/(V)xx(RT)/(pi)` `=(1.26)/(0.2)=(0.083xx300)/(2.57xx10^(-3))=61038"g mol"^(-1)`. |
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| 1385. |
Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases. |
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Answer» Vapour pressure of the solvent decrease in the presence of non-volatile solute (glucose) hence boiling point increases. |
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| 1386. |
The density of 10% by mass of KCI solution is 1.06g cm^(-3). Cuculate the molarity of the solution. |
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Answer» Calculaion of volume of the solution. Mass of soulution= 100g Density of solution=`1.06g cm^(3)` Volume of solution`=("Mass of solution")/("Density")=((100g))/((1.06g cm^(3)))=94.34cm^(3)` Calculation of molarity of the solution Molarity of solution (M)`= ("Mass of KCI"//"Molar mass of KCI")/("Volume of solution in" dm^(3))` Mass of KCI=10g, Molar mass of `KCI= 39+35.5=74.5" g mol"^(-1)` Volume of solution`=94.34cm^(3)=94.34cm^3=(94.34)/1000=0.0943 dm^(3)` Molarity(M)`= (10g//(74.74.5gmol^(-1)))/((0.0943" dm"^(3)))=1.42" mol dm"^(3)=1.42 M` |
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| 1387. |
Classify the following mixtures as Homogeneous and Heterogeneous Sugar solution, muddy water, mixture of water and kerosene, salt solution, air, smoke |
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| 1388. |
Tabulate the two differences of solution and colloid. |
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| 1389. |
Tabulate the components of solution given below.(saltwater, soda water, ornamental gold, air) |
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| 1390. |
Classify and tabulate the mixtures given below into solution, colloid, and suspension: Milk, fog, atmospheric air dilute acid, lime water, ink, smoke. |
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Answer» Solution:- Dilute acid, atmospheric air Colloid:- Milk, Smoke, Fog, Suspension:- Lime water |
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| 1391. |
When crystallizes of a solute are introduced into a saturated solution, the excess of solute crystallizes out and solution remains saturated (True/False). |
| Answer» Correct Answer - 1 | |
| 1392. |
What is the effect on solubility of a gas in a liquid when tempetature is decreased and gas pressure is increased ? |
| Answer» Correct Answer - Increases | |
| 1393. |
The solution which are isotonic with 6% (W/V) solution of urea are :A. 18% (W/V) solution of glucoseB. 0.5 M solution of `BaCI_(2)`C. 1M solution of sucreoseD. 1 M solution of scetic acid |
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Answer» Correct Answer - a,c (a,c) Osmotic pressure `(pi)=iCRT` `C_((Urea))=6g/100mL=60g/L=((60g))/((60"g mol"^(-1)))=1M` `ppi_((urea))=iCRT=RT(i=1)` `pi_((glucose))=18g/100mL=180g/L=((180g))/((180"g mol"^(-1)))=1M` `pi_("glucose")=iCRT=RT.I(i=1)` `(b)pi(BaCI_(2))=iCRT=(3)xx(0.5)RT=1.5RT` ltbtgt `(c) pi_(("sucrose"))=CRT=(1)RT=RT` `(d)pi_(("acetic acid"))=CRT=iCRT=2xx1RT=2RT` Thus (a) and (c) are isotonic with 6% solution of urea. |
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| 1394. |
A 6% of urea is isotonic withA. 1M solution of glucoseB. `0.05M` solution of glucoseC. 6% solution of glucoseD. 2.5% solution of glucose |
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Answer» Correct Answer - A Isotonic solution have the same molar concentration 6% urea means 6 g urea in 100 mL solution Moles of solute = 6/60 = 0.1 mol Volume of solution = 100 mL = 0.1 L `therefore` Molarity `= (0.1 mol)/(0.1 L)=1M` `therefore 6 %` urea solution is isotonic with 1M solution of glucose. |
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| 1395. |
10% solution of urea is isotonic with 6% solution of a non-volatile solute X.What is the molecular mass of solute X ?A. `"6 g mol"^(-1)`B. `"60 g mol"^(-1)`C. `"36 g mol"^(-1)`D. `"32 g mol"^(-1)` |
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Answer» Correct Answer - C No. of moles of urea = `10/60=1/6` Weight of solute , X=6 g No. of moles of X =`6/M` For isotonic solutions, `n_1=n_2` or `1/6=6/M` or `M=36 g mol^(-1)` |
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| 1396. |
The dissolving process is exothermic when :A. The energy released in solvation exceeds the energy used in breaking up solute-solute and solvent solvent interactions.B. The energy used in solvation exceeds the enrgy released in breaking up solute-solute and solvent solvent interactions.C. The energy released in solvation is about the same as the enrgy used in breaking up solute-solute and solvent-solvent interactions.D. The energy used in solvation is about the same as the energy used in breaking up solute-solute and solvent-solvent interactions. |
| Answer» Correct Answer - A | |
| 1397. |
The vapour pressure of water at `20^(@)C` is `17.54 mmHg`. What will be the vapour pressure of the water in the apparatus shown after the piston is lowered, decreasing the volume of the gas above the liquid to one half of its initial volume (assume temperature is constant). A. `8.77 mmHg`B. `17.54 mmHg`C. `35.08 mmHg`D. between `8.77` and `17.54 mmHg` |
| Answer» Correct Answer - B | |
| 1398. |
At `40^(@)C`, the vapour pressure os water is 55.3 mmHg. Calculate the vapour pressure at the same temperature over 10% aqueous solution of urea `[CO(NH_(2))_(2)]`. |
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Answer» Correct Answer - 53.48 mm Hg Mass of urea in 100 g of soution = 10 = 90 g `"Mass of water in the soltion"=100-10=90g` `"No. of moles of urea"=((10g))/((60"g mol"^(-1)))=5.0 mol` `"No. of moles of water" ((90g))/((18"g mol"^(-1)))=5.0 mol` `"Mole fraction of water"=((5.0 mol))/((5.0mol)+(0.17 mol))=(5.0)/(5.17)=0.967` `"Vapoure pressure of water over aquious solution "=P_("water")^(@)xxC_("water")=(55.3mm)xx0.967` =53.48 mm Hg. |
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| 1399. |
When 10.6 g of a nonvolatile substance is dissolved in 740 f of ether, its boiling point is raised `0.284^(@)"C"`. What is the molecular weight of the substance? Molal boiling point costant of ether is `2.11^(@)"C kg"//"mol"`. |
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Answer» Correct Answer - `"106 g/mol"` Solute-B, Ether-A `Delta"T"_("b")="K"_("b")xx("w"_("B")xx1000)/("m"_("B")xx740)` `0.284=211xx(10.6xx1000)/("m"_("B")xx740)` `"m"_("B")=106"gm mol"^(-1)` |
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| 1400. |
The vapour pressure of chlorobenzene and water at different temperature are `{:(t//^(@)C,90,100,110),(P^(@)(phiCI)//mmHg,204,289,402),(P^(@)(H_(2)O)//mmHg,526,760,1075):}` (A)At what pressure will `phiCl` steam-disillation at `90^(@)C`? (B)At what temperature will `phiCl` steam-dissillation under a total pressure of `800 mm Hg`? (C) How many grams of steam are required for distillation of `10.2g` of `phiCl` (i)at `90^(@)C` and (ii) under `800` torr total pressure? |
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Answer» Correct Answer - (A)`730mm Hg.` (B)`approx92^(@)C`, (C) `4.13gm,4.14gm` (A) At `90^(@)C`, Pressure at which `phiCl` steam `-` distil `=P^(@)(phiCl)+P^(@)(H_(2)O)` `=204+526=730 mm` of `Hg` (B) At `90^(@)C` Total `P_(T)=730mm Hg` At `100^(@)C` total `P_(T)=289+760=1049 mm Hg` so for `800 mm Hg`, temperature will lie in between `90^(@)-100^(@)C` Using extrapolation method. Temperature `=90+((800-730))/((1049-730))xx(100-90)=92.19^(@)C`. (C) (i) `(P_(phiCl)^(@))/(P_(H_(2)O)^(@))=(n_(phicl))/(n_(H_(2)O))` `(204)/(526)=(10xx18)/(112.5xxm)` `m=4.13gm` (ii) At `800` tprr `T=92^(@)C` For `phiCl,(p-204)/(92-90)=(289-204)/(100-90)` `P=572.8` For `H_(2)O,(p-526)/(92-90)=(760-526)/(100-90)` `P=572.8` `(10xx18)/(112.5xxm)=(221)/(572.8 rArrm=4.14gm` |
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