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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.
| 1451. |
A mixture of an organic liquid `A` and water distilled under one atmospheric pressure at `99.2^(@)C`. How many grams of steam will be condensed to obtain `1.0 g` of liquid `A` in the distillate? (Vapour pressure of water `99.2^(@)C` is `739 mm Hg`. Molecular weight of `A=123`) |
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Answer» Correct Answer - `5.15g` When the liquid `A-` water mixture distills, the pressure `=760 mm Hg`. `therefore` partial pressure of `A=760-739=21 mm Hg`. Mole fraction of `A` in the distillate `=(21)/(760)=0.02764` Mole fraction of water in the distillate `=(739)/(760)=0.9723` `therefore` mass of `A` in the distillate `=0.02764xx123=3.399g` Mass of steam condensed per `g` of liquid `A=(17.5)/(3.399)=5.15g`. |
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| 1452. |
Boiling point of a mixture of water and nitrobenzene is `99^(@)C`, the vapour pressure of water is `733 mm` of `Hg` and the atmospheric pressure is `760 n n` if `Hg`. The molecular weight of nitrobenzene is `123`. Find the ratio weights of the components of the distillate. |
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Answer» Correct Answer - `3.973` `(n_(W))/(n_(N))=(W_(w))/(18).(123)/(W_(N))=(733)/(33)rArr(W_(w))/(W_(N))=(18xx733)/(27xx123)=3.973` |
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| 1453. |
10 g of solute A and 20gm of solute B are both dissolved in 500ml water. The solution has the same osmotic pressure as 6.67 gm of A and 30 gm of B dissolved in the same amount of water at the same temperature. What is the ratio of molar masses of A and B? |
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Answer» Correct Answer - `M_(A)//M_(B)=0.33` `(("W"_("A"))/("m"_("A"))+("W"_("B"))/("m"_("B")))("RT")/("V")=(("W"_("A")^(1))/("m"_("A"))+("W"_("B")^(1))/("m"_("B")))("RT")/("V")` `(10)/("m"_("A"))+(20)/("m"_("B"))=(6.67)/("m"_("A"))+(30)/("m"_("B"))` `(3.33)/("m"_("A")=(10)/("m"_(B"))` `("M"_("A"))/("M"_("B"))=(3.33)/(10)` `("M"_("A"))/("M"_("B"))=0.33` |
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| 1454. |
The normal freezing point of nitrobenzene `(C_(6)H_(5)NO_(2))` is `278.82K`. `A 0.25` molal solution of a certain solute in nitrobenzene causes a freezing point depression of `2` degrees. Calculate the value of `K_(f)` for nitrobenzene. |
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Answer» Given Normal freezing point of nitrobenzene `(C_(6)H_(5)NO_(2))=278.82K` molality of solution `=0.25m` Freezing point depression `=2` degrees Asked Cryoscopic constant `=?` Formulae: `DeltaT_(f)=K_(f)xxm` Explanation: `DeltaT_(f)=` Freezing point depression `K_(f)=` cryoscopic constant `m=` molality of solution Substitution `&` Calculation `2=K_(f)xx0.25` `K_(f)=(2)/(0.25)=8K kg "mol"^(-1)` |
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| 1455. |
If `200ml` of `0.2M HgCl_(2)` solution is added to `800ml` of `0.5M Kl` (`100%` dissociated) solution. Assuming that the following complex formation taken place to `100%` extent. |
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Answer» `{:(HgCl_(2),+,4KI,rarr,K[HgI_(4)],+,2KCl.),(40,,400,,0,,0),(0,,400-160,,40,,80),(,,(240)/(1000),,(40)/(1000),,(80)/(1000)):}` `pi=(i_(1)C_(1)+i_(2)C_(2)+i_(3)C_(3))RT`. `=(0.24xx2+3xx0.04+0.08xx2)0.082xx300.=18.69` atm. Other method : `{:(HgCl_(2),+,4KI,hArr,K[HgI_(4)],+,2KCl.),(40,,400,,0,,0),(0,,400-160,,40,,80),(,,(240)/(1000),,(40)/(1000),,(80)/(1000)):}` `pi=(i_(1)C_(1)+C_(2)+i_(3)C_(3))RT`. `=(0.24xx2+xx0.04+0.08xx2)0.082xx300.=18.69` atm. |
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| 1456. |
If `10gm` of an unknown substance (non-electrolytic) is dissolved to make `500mL` of solution, then osmotic pressure at `300K` is observed to be `1.23` atm find molecular weight? |
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Answer» `1.23=(10xx1000)/(Mxx500)xx0.082xx300` `M=(20)/(1.23)xx(0.082)/(100)xx300 approx400gm//"mol"` |
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| 1457. |
If `6gm` of urea, `18gm` glucose `&` `34.2gm` surcose is dissolved to make `500mL` of a solution at `300K` calculate osmotic pressure? |
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Answer» molecular weight of urea `=62gm, "Glucose"=180gm, "Surcose"=342gm` `pi=Cxx0.082xx300` `pi=(0.3xx1000xx0.082xx300)/(500) rArr14.76` atm |
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| 1458. |
If same volume solution of different solute is used then what is order of (a) vapour pressure (b) moles solute (c) molar mass of solute. |
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Answer» `P_(A)prop1gm. :P_(A)-P_(B)prop0.5gm,P_(B)prop0.5gm.` `P_(C)-P_(B)prop1gm, P_(C)prop1.5gm.` `P_(C)gtP_(A)gtP_(B) , n_(c)ltn_(A)ltn_(B). , , M_(B)gtM_(A)gtM_(B).` |
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| 1459. |
The vapour pressure of benzene at `90^(@)C` is `1020` torr. A solution of `5g` of a solution in `58.5g` benzene has vapour pressure 990 torr. The molecular weight of the solute is? |
| Answer» `(P^(0)-P_(S))/(P_(S)) =(w xx M)/(m xx W) rArr (1020 -990)/(990) = (5 xx 78)/(m xx 58.5) rArr m = 220` | |
| 1460. |
Vapour pressure of `C CL_(4)` at `25^@C` is `143` mmHg 0.05g of a non-volatile solute (mol.wt.=`65`)is dissolved in `100ml C CL_(4)`. find the vapour pressure of the solution (density of `C CL_(4)=158g//cm^2`)A. `141.9 mmHg`B. `94.4mmHg`C. `-1.86^(@)C`D. `0.93^(@)C` |
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Answer» Correct Answer - A Wt. of `CCl_(4)=(153.8)/(154)approx1` moles of solute `=(0.5)/(65)=0.00769` Now `(P_(0)-P_(S))/(P_(S))=(n)/(N)` or `(143-P_(S))/(P_(S))=(0.00769)/(1)` so `P_(s)=141.9mm Hg`. |
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| 1461. |
The volume of `0.1M H_(2)SO_(4)` required to neutralise completely `40mL` of `0.2M NaOH` solution isA. `10mL`B. `40mL`C. `20mL`D. `80mL` |
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Answer» Correct Answer - B `2NaOH + H_(2)SO_(4) rarr Na_(2)SO_(4)+H_(2)O` `(M_(1)V_(1))/(n_(1)) = (M_(2)V_(2))/(n_(2))` `{:((NaOH),(H_(2)SO_(4)).),((0.2xx40)/(2)=,(0.1xxv)/(1)),(" "v=,40mL):}` |
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| 1462. |
Vapour pressure of `C CL_(4)` at `25^@C` is `143` mmHg 0.05g of a non-volatile solute (mol.wt.=`65`)is dissolved in `100ml C CL_(4)`. find the vapour pressure of the solution (density of `C CL_(4)=158g//cm^2`)A. `143.99`mmB. `94.39`mmC. `199.34`mmD. `14.197`mm |
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Answer» Correct Answer - D `(P^(@)-P_(S))/(P^(@))=(wxxM)/(mW)` `=143-1.03=141.97`mm. |
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| 1463. |
When `1.20g` of sulphur is melted with `15.00g` of naphthalene, the solution freezes at `77.2^(@)C`. What is the moalr mass of this from of sulphur. Data for Napthalene Melting point, m.p `80^(@)C` Freezing point depression constant, `k_(f) = 6.80^(@)Cm^(-1)`A. `180 g mol^(-1)`B. `190 g mol^(-1)`C. `260 g mol^(-1)`D. `450 g mol^(-1)` |
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Answer» Correct Answer - B A : Benzene B : Toluene `p = p_(A) + p_(B)` `p = p_(A)^(@) x_(A) + p_(B)^(@) x_(B)` `75 xx (1)/(2) + 22 xx (1)/(2)` `= 37.5 + 11 = 48.5` Mole fraction of benzene in vapour, `y_(A) = (p_(A))/(p) = (37.5)/(48) = 0.78` Similarly, mole fraction of toluene in vapour `y_(B) = 0.22` `:.` The vapour will contain higher precentage of benzene |
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| 1464. |
Which of the following concentration factors can be calculated if the mole fraction and density of an aqueous solution of `HCl` are known ?A. 1 onlyB. 3 onlyC. 1 and 2 onlyD. 1, 2 and 3 |
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Answer» Correct Answer - D `Delta T = K_(f) xx (w_(B) xx 1000)/(m_(B) xx w_(A))` `m_(B) = (K_(f))/(Delta T) xx (w_(B) xx 1000)/(w_(A))` `= (6.8)/(2.8) xx (1.2 xx 1000)/(15) = 190` |
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| 1465. |
Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to `4` kg for water to prevent it from freezing at `-6^(@)C` will be (`K_(f)` for water =`1.86 K kg mol^(-1)` and molar mass of ethylene glycol =`62 g mol^(-1)`)A. `800`gB. `204.30`gC. `400.00`gD. `304.60`g |
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Answer» Correct Answer - A `DeltaT_(f)` = Freezing point of `H_(2)O`-freezing point of ethylene glycol solution `=0-(-6^(@))=6^(@)` `K_(f)=1.86^(@)K kg mol^(-1)` `w_(1)`=Mass of ethylene glycol in grams `w_(2)`=Mass pf solvent `(H_(2)O)`in grams =`4000`g `m_(1)`=Molar mass of ethylene glycol=`62mol^(-1)` `DeltaT_(f)=(1000k_(f)w_(1))/(62xx4000)` `w_(1)=800g` |
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| 1466. |
Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to `4` kg for water to prevent it from freezing at `-6^(@)C` will be (`K_(f)` for water =`1.86 K kg mol^(-1)` and molar mass of ethylene glycol =`62 g mol^(-1)`)A. `304.60kg`B. `804.328g`C. `204.30g`D. `400.00g` |
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Answer» Correct Answer - B `DeltaT_(f) = k_(f) xx m = k_(f) xx (w_(2) xx 1000)/(w_(1) xx M_(2))` or `6 = (1.86 xx w_(2) xx 1000)/(4000 xx 62) :. W_(2) = 800g` |
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| 1467. |
Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to `4` kg for water to prevent it from freezing at `-6^(@)C` will be (`K_(f)` for water =`1.86 K kg mol^(-1)` and molar mass of ethylene glycol =`62 g mol^(-1)`)A. `400 g`B. `800 g`C. `304.60 g`D. `204.30 g` |
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Answer» Correct Answer - 2 We have `DeltaT_(f)=iK_(f)m` `=(iK_(f)W_("solute"))/(M_("solute") W_("solvent"))` Thus, `W_("solute")=(DeltaT_(f)M_("solute")W_("solvent"))/(iK_(f))` we are given `i=1` `K_(f)=1.86 K kg mol^(-1)` `DeltaT_(f)=T_(f^(0)-T_(f))=(0^(@)C)-(-6^(@)C)=6^(@)C= 6 K` `M_("solute")=62 g mol^(-1)` `W_("solvent")=4 kg` Substituting these results, we get `W_("solute")=((6 K)(62 g mol^(-1))(4 kg))/((1)(1.86 K kg mol^(-1))` `=800 g` |
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| 1468. |
Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to `4` kg for water to prevent it from freezing at `-6^(@)C` will be (`K_(f)` for water =`1.86 K kg mol^(-1)` and molar mass of ethylene glycol =`62 g mol^(-1)`)A. `400.00g`B. `304.60g`C. `804.32g`D. `204.30g` |
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Answer» Correct Answer - C `6 = 1.86 xx (w//62)/(4) rArr w = 800gm` |
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| 1469. |
Vapour pressure of `"CCl"_(4)` at `25^(@)C` is 143 mm Hg . 0.5 g of a non-volatile solute ( molar mass = `65 mol^(-1)`) is dissolved in 100 mL of `"CCl"_(4)` (density = 1.538g `mL^(-1)` ) Vapour pressure of solution is :A. `141.93mm`B. `94.39mm`C. `199.34mm`D. `143.99mm` |
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Answer» Correct Answer - A `W_(B) = 0.5g` `M_(B) = 65` amu `W_(A) = "vol" xx "density" = 100 xx 1.58 = 158g` `M_(A)(C CI_(4)) = 12 +35.5 xx 4 = 154` `(P^(@)-P_(s))/(P^(@)) = (W_(B))/(M_(B)).(M_(A))/(W_(A))` `P^(@) - P_(S) = (W_(B)M_(A))/(M_(B)W_(A)) xx P^(@)` `P_(S) = P^(@) - (W_(B)M_(A))/(M_(B)W_(A)) xx P^(@)` `= 143 - (0.5 xx 154)/(65 xx 158) xx 143` `= 143 - 1.03 = 141.97mm` |
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| 1470. |
Calculate the osmotic pressure of a solution obtained by mixing `100 cm^(3)` of `1.5%` solution of urea (mol. Mass=60) and `100 cm^(3)` of `3.42%` solution by cane sugar (mol. Mass = 342) at `20^(@)C`. (R=0.082 litre atm/deg/mole) |
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Answer» After mole total volume of the solution `=100 + 100=200 cm^(3)` Osmotic pressure due to the urea in the solution. `1.5 g` of urea which was present originally in `100 cm^(3)` is now present in `200 cm^(3)`,i.e., in the final solution. `Mw_(1)=Mw_("urea")=(W_(2) xx 1000)/(Mw_(2) xx vol. of sol.)=(1.5 xx 1000)/(60 xx 200)=0.125` `Mw_(2)=Mw_(sugar)=(3.42 xx 1000)/(342 xx 200)=0.05` `pi=(Mw_(1)+Mw_(2))RT` `=(0.125 + 0.05) xx0.82 xx 293 =4.2 atm` |
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| 1471. |
To neutralize completely 20 mL of 0.1M aqueous solution of phosphorus acid `(H_(3)PO_(3))` the volume of `0.1M` aqueous KOH solution required isA. `10mL`B. `20 mL`C. `40 mL`D. `60 mL` |
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Answer» Correct Answer - C `H_(3)PO_(3)` is a dibasic acid. `H_(3)PO_(3)+2KOH rarr K_(2)HPO_(3)+2H_(2)O` `(M_(A)V_(A))/(n_(A))=(M_(B)V_(B))/(n_(B))` `(0.1xx20)/(1) = (0.1xxV_(B))/(2)` `V_(B)=40 mL` |
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| 1472. |
Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to `4` kg for water to prevent it from freezing at `-6^(@)C` will be (`K_(f)` for water =`1.86 K kg mol^(-1)` and molar mass of ethylene glycol =`62 g mol^(-1)`)A. 304.6 gB. 804.32 gC. 204.3 gD. 400.00 g |
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Answer» Correct Answer - B `DeltaT=K_(f)xx(w_(B)xx1000)/(m_(B)xxw_(A))` `6=1.86xx(w_(B)xx1000)/(62xx4000)` `w_(B)=800g`] |
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| 1473. |
`6.02xx10^(20)` molecules of urea are present in 100 mL of its solution. The concentration of urea solution is:A. 0.02 MB. 0.01 MC. 0.001 MD. 0.1 M Glucose. |
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Answer» Correct Answer - b `"No. of moles of urea"=(6.02xx10^(20))/(6.02xx10^(23))` =0.001 mol. 100 mL of solution has moles = 0.001 1000 mL of solution has moles `=(0.001)/((10mL))xx(1000mL)=0.01 M.` |
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| 1474. |
To neutralise completely 20 mL of 0.1 M aqueous solution of phosphrous acid `(H_(3)PO_(3))`, the volume of 0.1 M aqueous KOH solution required is :A. 40 mLB. 20 mLC. 10 mLD. 60 mL |
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Answer» Correct Answer - a Since `H_(3)PO_(3)` is dibasic, 0.1 M solution =- 0.2 N solution. `underset(("Acid"))(N_(1)V_(1))=underset(("Base"))(N_(2)V_(2))` `(0.2 N)xx20 mL)=(0.1N)xxV_(2)` `V_(2)=((0.2N)xx(20mL))/((0.1N))=40 mL.` |
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| 1475. |
What is the osmotic pressure of the solution obtained by mixing `300 cm^(3)` of 2% (mass-volume) solution of urea with `300 cm^(3)` of 3.43 % solution of sucrose of `20^(@)C` ? `(R = 0.082 "L atm K"^(-1) "mol"^(-1))`A. 5 atmB. 5.2 atmC. 2.6 atmD. 4.5 atm |
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Answer» Correct Answer - B `piV=[((w_(B))/(m_(B)))_(urea)+((w_(B))/(m_(B)))_("sucrose")]RT` `pixx(600)/(1000)=[(6)/(60)+(3xx3.42)/(342)]0.082xx293` `pixx0.6=(0.1+0.03)xx0.082xx293` `pi=5.2`atm] |
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| 1476. |
A solution of sucrose (molar mass `342g mol^(-1)`has been produced by dissolving`68.5g`sucrose in `1000g`water .The freezing point of the solution obtained will be `(K_(f)` for `H_(2)O=1.86kgmol^(-1))`A. `-0.570^(@)C`B. `-0.372^(@)C`C. `-0.520^(@)C`D. `+0.372^(@)C` |
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Answer» Correct Answer - 2 We have `DeltaT_(f)=iK_(f)m` `i=1` for sucross as it neither associates nor dissociates `K_(f)=1.86 K kg mol^(-1)` `m=n_("sucrose")/(Kg_(H_(2)O))=(68.5 g//342 g mol^(-1))/(1 kg)` `=0.2 mol kg^(-1)` Substituting these result, we get `DeltaT_(f)=(1)(1.86 K kg mol^(-1))(0.2 mol kg^(-1))` `=0.372K or 0.372^(@)C` Freezing point of solution `=("Freezing point of water")-(DeltaT_(f))` `=(0.000^(@)C)-(0.372^(@)C)` `=-0.372^(@)C` |
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| 1477. |
A solution of sucrose (molar mass `342g mol^(-1)`has been produced by dissolving`68.5g`sucrose in `1000g`water .The freezing point of the solution obtained will be `(K_(f)` for `H_(2)O=1.86kgmol^(-1))`A. `-0.372^(@)C`B. `-0.520^(@)C`C. `+0.372^(@)C`D. `-0.570^(@)C` |
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Answer» Correct Answer - A `DeltaT_(f)="molarity" xxK_(f)` `(68.5xx1000)/(342xx1000)xx1.86=0.372` `:.T_(f)=0-0.732=-0.372^(@)C` |
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| 1478. |
A solution of sucrose (molar mass `= 342 g mol^(-1))` has been prepared by dissolving 68.5g of sucrose in 1000g of water. The freezing point of the solution obtained will be: `(K_(f)` for water `= 1.86 K kg mol^(-1))`A. `-0.520^(@)C`B. `+0.372^(@)C`C. `-0.570^(@)C`D. `-0.372^(@)C` |
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Answer» Correct Answer - D Depression in freezing point `DeltaT_(f)=K_(f)xxm` Molality `=(W_(B)xx1000)/(M_(B)xxW_(A))=(68.5xx1000)/(342xx1000)` `=(68.5)/(342)`m `:. DeltaT_(f)=1.86xx(68.5)/(342)=0.372^(@)C` `T_(f)=T_(f)^(@)-DeltaT_(f)=0.0-0.372=00.372^(@)C` . |
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| 1479. |
Benzene and toluene form nearly ideal solution. At `20^(@)C` the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at `20^(@)C` for a solution containing 78g of benzene and 46 g of toluene in torr is-A. `50`B. `25`C. `37.5`D. `53.5` |
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Answer» Correct Answer - A `P_(B)=P_(B)^(@)XB P_(B)^(@)=75` torr `X_(B)=(78//78)/((78//78)+(46//92))=(1)/(1+05)=(1)/(1.5)P_(B)=75xx(1)/(1.5)=50` torr. |
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| 1480. |
Benzene and toluene form nearly ideal solution. At `20^(@)C`, the vapour pressure of benzene is 75 torr and that of toluene ios 22 torr. The partial vapour pressure of benzene at `20^(@)C` for a salution containing 78 g of benzene and 46 g of toluene in torr is :A. 50B. 25C. 37.5D. 53.5 |
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Answer» Correct Answer - a `P_(B)^(@)=75"torr"` `X_(B)=(((78g))/((78gmol^(-1))))/(((78g mol^(-1)))/((78g mol^(-1)))+((46g))/((92gmol^(-1))))` `P_(B)^(@)=P_(B)^(@)X_(B)=75"torr"xx1/1.5=50 "torr"` |
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| 1481. |
Benzene and toluene form nearly ideal solution. At `20^(@)C` the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at `20^(@)C` for a solution containing 78g of benzene and 46 g of toluene in torr is-A. 25B. 50C. `53.5`D. `37.5` |
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Answer» Correct Answer - B Benzene: A, Tolvene: B `X_(A) = (1)/(1+0.5) = (1)/(3//2) = (2)/(3)` `P_(A) = P_(A)^(@) X_(A) = 75 xx (2)/(3) = 50` torr |
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| 1482. |
The volumes of `4N HCI` and `10 N HCI` required to make 1 litre of `6N HCI` areA. `0.75` litre of `10 N HCI` and `0.25` litre of `4N HCI`B. `0.25` litre of `4N HCI` and `0.75` litre of `10N HCI`C. `0.67` litre of `4N HCI` and `0.33` litre of `10N HCI`D. `0.80` litre of `4N HCI` and `0.20` litre of `10 N HCI` |
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Answer» Correct Answer - B Let volume of 4 N HCl used = `upsilon mL` therefore Volume of `10N HCl` used `= (1000 - upsilon) mL` `NxxV = N_(1)V_(1)+N_(2)V_(2)` `6xx1000=4xxupsilon+10(1000-upsilon)` `6upsilon = 10000-6000` `6 upsilon = 4000` `upsilon = (4000)/(6) = 666.7 mL` `=0.67 L` Vol. of 4 NHCl used `= 0.67 L` Vol. of 10 NHCl used `= 1-0.67 = 0.33 L` |
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| 1483. |
Benzene and toluene form nearly ideal solutuions. At `20^(@)C`, the vapour pressure of benzene is 75 torr and that of toulene is 22 torr. The partial vapour pressure of benzene at `20^(@)C` for a solution containing 78g of benzene and 46g of toluene in torr isA. 50B. 25C. 37.5D. 53.5 |
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Answer» Correct Answer - A Molar mass of `C_(6)H_(6)=78 g mol^(-1)` Molar mass of `C_(6)H_(5)-CH_(3)=92 g mol^(-1)` n(Benzene) = 78/78 = 1 n(Toluene) = 46/92 = 0.5 x(Benzene) = `(1)/(1+0.5)=(10)/(15)=(2)/(3)` p(Benzene) = x (Benzene) `xx P^(@)` (Benzene) `=(2)/(3)xx75 = 50` torr |
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| 1484. |
Which of the following colligative property can provide molar mass of proteins (or polymers or colloids) with greatest precision?A. Relative lowering of vapour pressureB. Elevation of boiling pointC. Depression in freezing pointD. Osmotic pressure |
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Answer» Correct Answer - D Molar mass of macro-molecules (proteins, polymers of colloids) can be found quite accurately by finding the osmotic pressure as magnitude of this colligative property is comparatively quite large. |
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| 1485. |
Which of the following colligative property can provide molar mass of proteins (or polymers or colloids) with greatest precision?A. Osmotic pressureB. Elevation in boiling pointC. Depression in freezing pointD. Relative lowing of vapor pressure |
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Answer» Correct Answer - 1 The most important use of colligative properties in the laboratory is for determining the molecular mass of an unknown non-volative substance. Any of the four colligative properties can be used, but the most accurate values are obtained from osmotic-pressure measurements because the magnitude of the osmosis effect is so great. For example, a solution of `0.200 M` glucose in water at `300 K` will give an osmotic-pressure reading of `374.2 mm Hg`, a value that can easily be read to four significant figures. However, the same solution will lower the freezing point by only `0.04^(@)C`, a value that can be read to only one significant figure. |
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| 1486. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day to day life.one example is the use of ethylene glycol and water mixture as antifreezing liquid in the radiators of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in water is 0.9. Given, freezing point depression constant of water `(K_(f) "water") = 1.86 Kg mol^(-1)`. Freezing point depression constant of ethanol`K_(f)` (ethanol) `= 2.0 K kg mol^(-1)` Boiling point elevation constant of water `K_(b)` (water) = `0.52 K kg mol^(-1)` Boiling point elevation constant of ethanol `K_(b)` (ethanol) `= 1.2 K kg mol^(-1)` Standard freezing point of water `= 273 K` Standard freezing point of ethanol `= 155.7 K` Standard boiling point of water `= 373 K` Standard boiling point of ethanol `= 351.5 K` Vapour pressure of pure water `= 32.8 mm Hg` Vapoure pressure of pure ethanol `= 40 mm Hg` Molecular weight of water `= 18 g mol^(-1)` Molecular weight of ethanol `= 46 g mol^(-1)` In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution M isA. `39.3 mm Hg`B. `36.0 mm Hg`C. `29.5 mm Hg`D. `28.8 mm Hg` |
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Answer» Correct Answer - B Total vapour pressure, `p = p_(A)^(@)xxA` `p= 40 xx 0.9= 36` mm Hg Since it is given that solute is taken as non volatile, water does not contribute to the total vapour pressure. |
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| 1487. |
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day to day life.one example is the use of ethylene glycol and water mixture as antifreezing liquid in the radiators of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in water is 0.9. Given, freezing point depression constant of water `(K_(f) "water") = 1.86 Kg mol^(-1)`. Freezing point depression constant of ethanol`K_(f)` (ethanol) `= 2.0 K kg mol^(-1)` Boiling point elevation constant of water `K_(b)` (water) = `0.52 K kg mol^(-1)` Boiling point elevation constant of ethanol `K_(b)` (ethanol) `= 1.2 K kg mol^(-1)` Standard freezing point of water `= 273 K` Standard freezing point of ethanol `= 155.7 K` Standard boiling point of water `= 373 K` Standard boiling point of ethanol `= 351.5 K` Vapour pressure of pure water `= 32.8 mm Hg` Vapoure pressure of pure ethanol `= 40 mm Hg` Molecular weight of water `= 18 g mol^(-1)` Molecular weight of ethanol `= 46 g mol^(-1)` In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of solution M isA. `268.7 K`B. `268.5 K`C. `234.2 K`D. `150.9 K` |
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Answer» Correct Answer - D Solution M is mixture of ethanol and water, Mole fraction of ethanol is 0.9 (Solvent is `C_(2)H_(5)OH`) Mole fraction water is 0.1 (`H_(2)O` is solute) Molality of `H_(2)O=(n_(2)xx1000)/(n_(1)M_(1))` `=(0.1xx1000)/(0.9xx46)=2.415` m `DeltaT_(f)=K_(f)m=2xx2.2415=4.83` Freezing point of solution `= 155.7 - 4.83 = 150.87 K` Total vapour pressure, `p=p_(A)^(@)xx A`. |
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| 1488. |
A 5.2 molal aqueous solution of methyl alcohol, `CH_(3)OH`. What is the mole fraction of methyl alcohol in solution?A. `0.190`B. `0.080`C. `0.050`D. `0.100` |
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Answer» Correct Answer - D Moles of `CH_(3)OH = 5.2` Mass of `CH_(3)OH=5.2xx32=166.4g` Mass of water `=1000-166.4=833.6` Moles of water `= (833.6)/(18)=46.3` Mole fraction of `CH_(3)OH=(5.2)/(46.3+5.2)` `=(5.2)/(51.5)=0.1`. |
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| 1489. |
Dissolving `120g` of urea (mol wt `= 60)` in 1000g of water gave a solution of density 1.15 g/mL. The molarity of the solution isA. `1.78M`B. `2.00M`C. `2.05M`D. `2.22M` |
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Answer» Correct Answer - C Mass of urea = 120g Mass of water = 1000g Mass of solutuion = 1120g Density of solution = 1.15 g/mL Volume of the solution `= ("mass")/("density")=(1120)/(1.15)` `= 973.9 mL = 0.974 L` Moles of urea `= (120)/(60)=2` Molarity `= ("2 mol")/(0.974L)=0.05 M` |
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| 1490. |
Assuming each salt to be `90%` dissociated which of the following will have the highest osmotic pressure?A. Decinormal `Al_(2)(SO_(4))_(3)`B. Decinormal `BaCl_(2)`C. Decinormal `Na_(2)SO_(4)`D. A solution obtained by mixing equal volumes of (b) and (c) and filtering |
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Answer» Correct Answer - A `Al_(2)(SO_(4))_(3)` furnishes maximum number of ions. |
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| 1491. |
The molal boiling point constant of water is `0.573^(@)C kg "mole"^(-1)`. When `0.1` mole of glucose is dissolved in `1000g` of water, the solution boils under atmospheric pressure at:A. `100.513^(@)C`B. `100.0573^(@)C`C. `100.256^(@)C`D. `101.025^(@)C` |
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Answer» Correct Answer - B `T - 100 = 0.573 xx 0.1` `T = 100.0573` |
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| 1492. |
The relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute. This statement was given byA. RaoultB. HenryC. JouleD. Dalton |
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Answer» Correct Answer - A This is Raoult,s law. |
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| 1493. |
STATEMENT-1 : Relative lowering of vapour pressure is equal to mole fraction of the solute. and STATEMENT-2 : Relative lowering of vapour pressure is a colligative property.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-2B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - 2 |
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| 1494. |
Two liquids X and Y on mixing gives a warm solution. The solution is ……(a) ideal (b) non-ideal and shows positive deviation from Raoults law (c) ideal and shows negative deviation from Raoults Law (d) non – ideal and shows negative deviation from Raoults Law |
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Answer» (d) non – ideal and shows negative deviation from Raoults Law ∆Hmix is negative and show negative deviation from Raoults law. |
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| 1495. |
Azeotropic mixtures areA. Constant boiling point mixture without changing the composition.B. Those which boil at different temperatures.C. Mixtures of two solids.D. None of the above |
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Answer» Correct Answer - A The difinition of azeotropic mixture. |
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| 1496. |
The vapour pressure of benzene at`80^(@)C` is lowered by `10mm` by dissoving `2g`of a non-volatile substance in`78g`of benzene .The vapour pressure of pure benzene at `80^(@)C`is `750mm`.The molecular weight of the substance will be:A. `15`B. `150`C. `1500`D. `148` |
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Answer» Correct Answer - D `(P_(0)-P_(s))/(P_(s))=(wxxM)/(mxxW)` `(10)/((750-10))=(2xx78)/(mxx78)` `:. m= 148`, Note : [`m` comes `150` if formula `(P_(0)_p_(s))/(P_(o))=(wxxM)/(mxxW)` is used. But this is only for dilute solutions]. |
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| 1497. |
Assertion :Molar heat of vaporisation of water is greater than benzene. Reason:Molar heat of vaporisation is the amount of heat required to vaporise one mole of liquid at constant temperature.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 reason both are false. |
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Answer» Correct Answer - B We know that heat of vaporisation of water is `100^(@)C`is `40.6kJ`and that of benzene is `31kJ`at `80^(@)C`. The amount of heat required to vaporise one mole of liquid at constant temperature is known as heat of vapourisation therefore both assertion and reason are true but reason is not the correct explanation of assertion. |
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| 1498. |
Asssertion : Azeotropic mixtures are formed only by non-ideal solutions and they may have boiling points either greater than both the components or less than both the compounds. Reason : The composition of the vapour phase is same as that of the liquid phase of an azeotropic mixture.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 reason both are false. |
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Answer» Correct Answer - B Non-ideal solutions with positive deviation i.e,having more vapour pressure than expected ,boil at lower temperature while those with negative devaition boil at higher temperature than those of the components. |
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| 1499. |
The freezing point of solution containing `0.2 g` of acetic acid in `20.0 g` of benzene is lowered by `0.45^(@)C`. Calculate the degree of association of acetic acid in benzene. `(K_(f)=5.12 K^(@) mol^(-1) kg^(-1))` |
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Answer» Correct Answer - 0.945 Acetic acid -B, Benzene-A For acetic acid,` "n=2"` `Delta"T"_("f")="i"xx("K"_("f")xx"W"_("B")xx1000)/("m"_("B")xx"W"_("A")("g"))" "0.45xx"i"xx(5.12xx0.2xx1000)/(60xx20)` `"i=0.527"` `1-(alpha)/(2)=1` `1-(alpha)/(2)=0.527` `alpha=0.945` `alpha=94.5%` |
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| 1500. |
Select correct statements :A. Osmosis, like all colligative properties, results from an increase in entropy as pure solvent passes through the membrane and mixes with the solutionB. Desalination of sea-water is done by reverse osmosisC. Both are correct statementsD. fNone is correct statement |
| Answer» Correct Answer - C | |