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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.
| 1951. |
Calculatte the molal elevation constant of water if molar enthalpy of vapourisation of water at 373K is 40.585 kJ `mol^(-1)`. |
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Answer» `K_(b)=(MR(T_(b)^(@))^(2))/((DeltaH_(VAP))),DeltaH_((VAP))=40.585 kJ mol^(-1)=40585 J mol^(-1)` `R=8.314 jK^(-1)mol^(-1),T_(b)^(@)=373K,18 g mol^(-01)=0.018 kg mol^(-1)` `K_(b)=((0.018 kg mol^(-1))xx(8.314JK^(-1)mil^(-1)xx(373K)^(2)))/((40585 J mol^(-1)))` = 0.513 K kg `mol^(-1)`=0.513 K/m |
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| 1952. |
The boling point of benzone is 353.23K. When 1.80 g of a non-valatioe solute is mixed in 90 g of benzene, the boling point id raised to 354.11 Calculate the molar mass of the solute. Given that `K_(b)` benzene is 2.53 K kg `mol^(-1)` |
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Answer» `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))` `W_(B)=1.80 kg,k_(b)=2.53K Kg mol^(-1)` `DeltaT_(b)=(354.11-353.23)=0.88 K` `M_(B)=((1.80)xx(2.53 kg mol^(-1)))/((0.88)xx(0.090 kg))=9.5 g mol^(-1)` |
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| 1953. |
A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boling point of `80.31^(@)C`. Determine the molar mass of this compound (B.P. Of pure benzene = `80.10^(@)C`, and `K_(b)` for benzene= `2.53^(@)K kg mol^(-1))` |
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Answer» `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))` `K_(b)=2.53 K kg mol^(-1),W_(B)=1.25g, W_(A)=99.0g=0.099 kg, DeltaT_(b)=(80.31-80.10)=0.21^(@)=0.21K` `M_(B)=((2.53K kgmol^(-1))xx(1.25g))/((0.21K)xx(0.099 kg))=152.12 g mol^(-1)` |
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| 1954. |
A solution prepared by dissolving `1.25g` of oil of winter green (methyl sallicylate) in `99.0g` of benzene has a boiling point of `80.31^(@)C`. Determine the molar mass of this compound. (`B.P.` of pure benzene `=80.10^(@)C` and `K_(b)` for benzene `=2.53^(@)C kg "mol".1`) |
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Answer» Correct Answer - Mass of solute `(W_(B))=1.25g` Mass of solvent `(W_(A))=99g` Elevation in boiling point `(DeltaTb)=8031-80.10^(@)C` `K_(b)=2.53^(@)Ckg "mol"^(-1)` Now, `DeltaT_(b)=K_(b)(W_(b)xx100)/(W_(A)xxM_(B))` `0.21=(2.53xx1.25xx1000)/(99xxM_(B))` or `M_(B)=(2.53xx1.25xx1000)/(99xx0.21)` ir `M_(B)=152.116g` |
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| 1955. |
How many ml of 0.1 HCl is required to react completely with 1.0g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equi-molar amounts of both ? |
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Answer» Given 1gm mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` Let the mass of `Na_(2)CO_(3)=" a gms"` Mass of `"NaHCO"_(3)=(1-a)" gms"` Number of moles of `"Na"_(2)"CO"_(3)=("wt")/("GMW")=(a)/(106)` Number of moles of `"NaHCO"_(3)=("wt")/("GMW")=(1-a)/(84)` Given that the mixture contains Equi molar amounts of `"Na"_(2)"CO"_(3)` and `"NaHCO"_(3)` `:. (a)/(106)=(1-a)/(84)` `84" a"=106-106" a"` `190"a"=106` a = 0.558 gms `:." Weight of "Na_(2)CO_(3)=0.558" gms"` Weight of `"NaHCO"_(3)=1-0.558=0.442" gms"` `"Na"_(2)"CO"_(3)+2" HCl"to2" NaCl"+"H"_(2)"O"+"CO"_(2)` 106 gms - 73 gms `0.558"g"-?=(73xx0.558)/(106)=0.384" gms"` `"NaHCO"_(3)+"HCl" to" NaCl"+"H"_(2)"O"+"CO"_(2)` 84 gms - 36.5 gms 0.442gms - ? `=(36.5xx0.442)/(84)=0.1928` `:." The weight of HCl required "=0.384+0.192=0.576" gms"` Molarity (M) `=("Wt")/("GMWt")xx(1000)/("V"(ml))` `0.1=(0.576)/(36.5)xx(1000)/("V")` `V=(0.576xx1000)/(36.5xx0.1)=(576)/(3.65)=157.80" ml"` |
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| 1956. |
A `5.25%` solution of a substance is isotonic with a `1.5%` solution of urea (molar mass `= 60g mol^(-1)`) in the same solvent. If the densities of both the solutions are assumed to be equal to `1.0 g cm^(-3)`, molar mass of the substance will be:A. `210.0 g mol^(-1)`B. `90.0 g mol^(-1)`C. `115.0 g mol^(-1)`D. `105.0` |
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Answer» Correct Answer - A Isotonic solutions have the same molar concentration. Hence `(52.5"g L"^(-1))/(M)=("15 g L"^(-1))/("60 g mol"^(-1))` (100g solution = 100 mL as `d = 1 g mL^(-1)`) `M=(52.5xx60)/(15)=210"g mol"^(-1)`. |
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| 1957. |
A `5.25%` solution of a substance is isotonic with a `1.5%` solution of urea (molar mass `= 60g mol^(-1)`) in the same solvent. If the densities of both the solutions are assumed to be equal to `1.0 g cm^(-3)`, molar mass of the substance will be:A. `105.0g "mol"^(-1)`B. `210.0g "mol"^(-1)`C. `90.0g "mol"^(-1)`D. `15.0g "mol"^(-1)` |
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Answer» Correct Answer - B Isotonic solution have same osmotic pressure. `pi_(1)=C_(1)RT,pi_(2)=C_(2)RT` For isotonic solution, `pi_(1)=pi_(2)` `therefore C_(1)=C_(2)`. or `(1.5//60)/(V)=(5.25//M)/(V)` [where `M=` molecular weight of the substance] or `(1.5)/(60)=(5.25)/(M)=210`. |
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| 1958. |
A solution is obtained by mixing 300g of `25%` solution and 400g of `40%` solution by mass. Calculate the mass percentage of the resulting solution. |
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Answer» Given a solution is obtained by mixing 300 gms of `25%` solution and 400 gms of `40%` solution by mass. Weight of solute in `1^("st")` solution `=300xx(25)/(100)=75" gms"` Weight of solute in `2^(nd)` solution `=400xx(40)/(100)=160" gms"` Total weight of solute `=75+160=235" gms"` Total weight of solution `=300+400=700" gms"` Mass % of solute in resulting solution `=(235)/(700)xx100=33.5%` |
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| 1959. |
The mass of ascorbic acid `(C_(6)H_(8)O_(6))` to be dissolved in 100g of acetic acid to lower its freezing point by `1.5^(@)C` in g is : (`K_(f)` for acetic acid is `"4.0 K kg mol"^(-1)`)A. 17.6B. 8.8C. 6.6D. 13.2 |
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Answer» Correct Answer - C `DeltaT=k_(f)xx(w_(f)xx1000)/(m_(B)xxw_(A))` `1.5=4xx(w_(B)xx1000)/(176xx100)` `w_(B)=6.6g`] |
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| 1960. |
A solution containing `24g` of a non-electrolyte per `kg` of water starts to freeze at `-0.75^(@)C`. The molar mass of the solute is `60g "mol"^(-1)`. Calculate the molal depression constnat for water. If the solution is cooled to `-1^(@)C`. How much of ice would separate ? |
| Answer» Correct Answer - `1.875.250gm` | |
| 1961. |
A `0.001` molal solution of a complex represented as `Pt(NH_(3))_(4)Cl_(4)` in water had freezing point depression of `0.0054^(@)C`. Given `K_(f)` for `H_(2)O=1.86 K m^(-1)`. Assuming `100%` ionization of the complex, write the ionization nature and formula or complex.A. `["MA"_(8)]`B. `["MA"_(7)]"A"`C. `["MA"_(6)]"A"_(2)`D. `["MA"_(5)]"A"_(3)` |
| Answer» Correct Answer - C | |
| 1962. |
A `0.001` molal solution of `[Pt(NH_(3))_(4)Cl_(4)]` in water had freezing point depression of `0.0054^(@)C`. If `K_(f)` for water is `1.80`, calculating the number f `Cl^(-)` ions furnished.A. `[Pt(NH_(3))_(4)Cl_(3)]Cl`B. `[Pt(NH_(3))_(4)Cl_(3)]Cl_(2)`C. `[Pt(NH_(3))_(4)Cl_(3)]Cl_(3)`D. `[Pt(NH_(3))_(4)Cl_(3)]Cl_(4)]` |
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Answer» Correct Answer - B `DeltaT_(f)=imk_(f),0.0054=ixx1.8xx0.001` `i=3`so it is `[Pt(NH_(3))_(4)Cl]Cl_(2)` |
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| 1963. |
At `300 K, 40 mL` of `O_(3)(g)` dissolves in `100g` of water at `1.0atm`. What mass of ozone dissolved in `400g ` of water at a pressure of `4.0atm` at `300 K` ? |
| Answer» Correct Answer - `4.8g` | |
| 1964. |
The osmotic pressure of a solution (density is 1 g `mL^(-1))` containing `3 g` of glucose (molecular weight =180) in `60 g` of water at `15^(@)C` isA. `0.34 atm`B. `0.65 atm`C. `6.25 atm`D. `5.57 atm` |
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Answer» Correct Answer - C `pi xx 63/1000=3/180 xx 0.0821 xx 288` `:. pi = 6.25` atm `:. V_("solution")= "Weight of solution" xx "Density"` |
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| 1965. |
Dry air was successively passed through a solution of `5g` solute in `80g` water and then through pure water. The loss in weight of solution was `2.5g` and that of pure water was `0.04g`. What is mol. wt. of solute ? |
| Answer» Correct Answer - `m=70.31` | |
| 1966. |
Dry air was passed successively through solution of `5g` of a solute in `180g` of water and then through pure water. The loss in weight of solution was `2.50 g` and that of pure solvent `0.04g`. The molecualr weight of the solute is:A. 31.25B. 3.125C. 312.5D. None |
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Answer» Correct Answer - A `P^(@)-P_(S) prop ` loss in weight of water chamber. and `P_(S) prop ` loss in weight of solution chamber. `(P^(@)-P_(S))/-P_(S) = n_(2)/n_(1) =(W_(2) xx Mw_(1))/(Mw_(2) xx W_(1))` or `0.04/2.50 = (5 xx 18)/(Mw_(2) xx 180)` `:. Mw_(2) =31.25` |
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| 1967. |
The osmotic pressure of blood is `7.65`atm.at`310K`An aqueous solution of glucose which is isotonic with blood has the percentage (wt./volume)A. `5.41%`B. `3.54%`C. `4.53%`D. `53.4%` |
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Answer» Correct Answer - A `piV=nST` for glucose and blood , if isotonic `pi_("glucose")=pi_("blood")` Thus `7.65xxV=(w)/(180)xx0.0821xx310` `:.(w)/(V)=54.1g//L` or `5.41%` |
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| 1968. |
A solution containing`8,6` urea in one litre was found to be isotonic with a `5%`(wt./vot.)solution of an organic non-volatile solute. The molecular weight of latter is:A. `348.9`B. `34.89`C. `3489`D. `861.2` |
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Answer» Correct Answer - A For two non-electrolytic solutions if isotonic`C_(1)=C_(2)` `:.(8.6)/(60xx1)=(5xx1000)/(mxx100):.m=348.9` |
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| 1969. |
The solution containing `4.0gm` of a polyvinyl chloride polymer in `1` litre dioxane was found to have an osmotic pressure `6.0xx10^(-4)`atmosphere at `300K`,the value of R used is `0.082`litre atmosphere`"mole"^(-1)K^(-1)`.The molecular mass of the polymer was found to beA. `3.0xx10^(2)`B. `1.6xx10^(5)`C. `5.6xx10^(4)`D. `6.4xx10^(2)` |
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Answer» Correct Answer - B `!=V=(w)/(m)RT` `:.6xx10^(-4)xx1=(4)/(m)xx0.0821xx300rArrm=1.64xx10^(5)` |
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| 1970. |
A solution containing `10 g per dm^(3)` of urea (mol.wt. `= 60 g mol^(-1)`) is isotonic with a `5%` ( mass//vol.) of a non-volatile solute. The molecular mass of non-volatile solute is:A. `"250 g mol"^(-1)`B. `"300 g mol"^(-1)`C. `"350 g mol"^(-1)`D. `"200 g mol"^(-1)` |
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Answer» Correct Answer - B `pi_(1)` (urea) = `pi_(2)` (unknown solute) `C_(1)` (urea) = `C_(2)` (unknown solute) `[(w_(B)xx1000)/(m_(B)xxV)]_("urea")=[(w_(B)xx1000)/(m_(B)xxV)]_("unknown solute")` `(10xx1000)/(60xx1000)=(5xx1000)/(m_(B)xx100)` `m_(B)=300 "g mol"^(-1)`] |
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| 1971. |
The temperature at which 10% aqeous solution `((W)/(V))` of glucose will exhibit the osmotic pressure of 16.4 atm, is : `(R = 0.082 "dm"^(2) "atm K"^(-1) "mol"^(-1))`A. `360^(@)C`B. 180 KC. 90 KD. 360 K |
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Answer» Correct Answer - D `pi=nRT` `piV=(W)/(m)=RT` `16.4xx((100)/(1000))=(10)/(180)xx0.082xxT` `:. T=360K`] |
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| 1972. |
The osomotic pressure of a solution at `0^(@)C` is `4 atm`. What will be its osmotic pressure at `546 K` under similar conditions? a.`4 atm`, b.`9 atm`,c.`8 atm`, d.`6 atm`A. 4 atmB. 2 atmC. 8 atmD. 1 atm |
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Answer» Correct Answer - C `(pi_(1))/(pi_(2))=(CRT_(1))/(CRT_(2))` `(pi_(1))/(pi_(2))=(T_(1))/(T_(2))` `(4)/(pi_(2))=(273)/(546)` `pi_(2)=8` atm] |
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| 1973. |
3g urea is dissolved in 45g of water. The relative lowering of vapour pressure is :A. 0.05B. 0.04C. 0.02D. 0.01 |
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Answer» Correct Answer - C `n_(B)=(3)/(60)=0.05,n_(A)=(45)/(18)=2.5` `(Deltap)/(p_(0))=x_(B)=(0.05)/(2.5+0.05)=0.0196=0.02`] |
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| 1974. |
6g of urea is dissolved in 90g of water. The relative lowering of vapour pressure is equal toA. 0.02B. 0.05C. 0.1D. 0.04 |
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Answer» Correct Answer - A `(DeltaP)/(p^(@)) = x_(B) ~~ (6 xx 18)/(60 xx 90) = 0.02` |
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| 1975. |
An aqueous solution of glucose was prepared by dissolving 18g of glucose in 90g of water. The relative lowering in vapour pressure isA. 0.02B. 1.1C. 20D. 196 |
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Answer» Correct Answer - A `(DeltaP)/(p^(@)) = (18 xx 18)/(180 xx 90) = 0.02` |
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| 1976. |
The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1g of urea in the same quantity of water is :A. 1gB. 3gC. 6gD. 18g |
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Answer» Correct Answer - B `(Deltap)_("glucose")=(Deltap)_("urea")` `(x_(B))_("glucose")=(x_(B))_("urea")` i.e., `((n_(B))/(n_(A)))_("glucose")=((n_(B))/(n_(A)))_("urea")` `(w_(B))/(50)xx(18)/(180)=(1xx18)/(50xx60)` `w_(B)=3g`] |
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| 1977. |
An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following. Depression in freezing point of solutionA. `0.68`B. `0.43`C. `0.5989`D. `0.326` |
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Answer» Correct Answer - C Depression in freezing point `(DeltaT_(f))` `DeltaT_(f) = K_(f)m` `=1.86 xx 0.322` `=0.5989 K` |
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| 1978. |
An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following. Boiling point of the solution isA. `300.73 K`B. `373.165 K`C. `400 K`D. `273.15 K` |
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Answer» Correct Answer - B Boiling point of solution `DeltaT_(f)=K_(b) xx m` `K_(b)=0.512 K kg mol^(-1)`,`m=0.322 m` ,`DeltaT_(b)=?` `DeltaT_(b)=0.512 xx 0.322 = 0.165^(@)` Boiling point of solution =`373 + 0.165 =373.165 K` |
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| 1979. |
An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following. Molality of the solution isA. `0.322`B. `0.222`C. `0.413`D. `0.5` |
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Answer» Correct Answer - A Molarity of solution `DeltaT_(f)=K_(f) xx m rArr =(DeltaT_(f))/(K_(f))` `DeltaT_(f)=273-272.4 = 0.6K`,`K_(f)=0.86 K kg mol^(-1)` `:.m=0.6/1.86=0.322` |
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| 1980. |
An aqueous solution of sodium chloride freezes below 273 K. Explain the lowering in freezing points of water with the help of a suitable diagram. 1 |
| Answer» Correct Answer - When `NaCl` is dissolved in water, the vapour pressure of solution becomes less and at a lower temperature vapour pressure of solid solvent and liquid solvent becomes equal, i.e., lowering in freezing point takes place. | |
| 1981. |
In an Adsorption experiment a graph between log `x//m` versus log P was found to be linear with a slope of `45^(@)` the intercept of the log `x//m` was found to be `0.3010`. Calculate the amount of gas adsorbed per gm of charcoal under a pressure of `0.6` bar. |
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Answer» Correct Answer - `1.2` According to frendulish equation `x/m=KxxP^(1//n)` `"log" x/m=1/n "log" P+log K` compare with straight line equation ltbtgt `1/n= tan `45^(@), n=1` intercept `loh K=0.03010` `K=2` Put the value of K, P & n in freundlich equation `x/m=KxxP^(1//n)=2xx(0.6)^(1)=1.2` |
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| 1982. |
Assertion: A quious gold colloidal solution is red in colour. Reason: The colour arises due to scattering of light by colloidal gold particles.A. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-6B. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-7C. STATEMENT-1 is true, STATEMENT-2 is falseD. STATEMENT-1 is false, STATEMENT-2 is true |
| Answer» Correct Answer - A | |
| 1983. |
Assertion: Tetraethyl lead minimizes the knocking effect when mixed with petrol. Reason: Because tetraethyl lead acts as a `-Ve catalyst.A. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-13B. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-14C. STATEMENT-1 is true, STATEMENT-2 is falseD. STATEMENT-1 is false, STATEMENT-2 is true |
| Answer» Correct Answer - A | |
| 1984. |
Assertion: Physisorption of molecules occures on surface only. Reason: in this process, the bonds of the adsorbed molecules are broken.A. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-11B. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-12C. STATEMENT-1 is true, STATEMENT-2 is falseD. STATEMENT-1 is false, STATEMENT-2 is true |
| Answer» Correct Answer - C | |
| 1985. |
How much of NaOH is reuired to neutralise 1500 `cm^(3)` of 0.1 N HCl (Na=23)?A. 60gB. 6gC. 4gD. 40g |
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Answer» Correct Answer - B `1500 cm^(3)` of `0.1N HCI -= 1500 cm^(3)` of `0.1N NaOH` `-= 1500 cm^(3)` of `0.1M NaOH` `=((1500)/(1000)L)xx (0.1 mol L^(-1)) = 0.15` mol `= 0.15` mol `xx 40 g mol^(-1) = 6g` |
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| 1986. |
`FeCl_(3)` on reaction with `K_(4)[Fe(CN)_(6)]` in aqueous solution gives blue colour. These are separated by a semi-permeable membrane `AB` as shown. Due to osmosis, there is A. Blue colour formation in side `X`.B. Blue colour formation in side `Y`.C. Blue colour formation in both of side `X` and `Y`.D. No blue colour formation. |
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Answer» Correct Answer - D Osmosis of water and not of ion takes place. |
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| 1987. |
The osmotic pressure of a dilute solution is given byA. `P=P_(0)xxN_(1)`B. `piV=nRT`C. `DeltaP=P_(0)N_(2)`D. `(DeltaP)/(P^(@))=(P^(@)-P_(s))/P^(@)` |
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Answer» Correct Answer - B `piV = nRT rArr pi =n/V(RT) rArr pi =CRT` |
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| 1988. |
According to Boyle-Vant Hoff law for solutions, the osmotic pressure of a dilute solutions isA. equal to its volumeB. directly proportional to its volumeC. inversely proportional to its volumeD. none of the above |
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Answer» Correct Answer - C `pi = (nRT)/(V)` |
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| 1989. |
The solutions having same osmotic pressure w.r.t. same semi-permeable membrane are calledA. Equivalent solutionsB. Ideal solutionsC. Equimolar solutionsD. Isotonic solutions |
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Answer» Correct Answer - D Solutions with same osmotic pressure are called isotonic solutions. Two equimolar solutions may have different osmotic pressure if one or both of solutes associates of dissociates to different extents. |
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| 1990. |
The osmotic pressure of a dilute solution is directly proportional to theA. Diffusion rate of the soluteB. Ionic concentrationC. Boiling pointD. Flow of solvent form a concentrated solution |
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Answer» Correct Answer - B `piV=nRT rArr pi=n/VRT rArr pi =CRT` |
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| 1991. |
Each pair forms ideal solution exceptA. `C_(2)H_(5)Br` and `C_(2)H_(5)I`B. `C_(2)H_(5)Cl` and `C_(2)H_(5)Br`C. `C_(6)H_(6)` and `C_(6)H_(5)CH_(3)`D. `C_(2)H_(6)I` and `C_(2)H_(5)OH` |
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Answer» Correct Answer - D `C_(2)H_(5)OH` show H-bonding as well as polarity both. |
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| 1992. |
What will happen if pressure greater than the osmotic pressure is applied on the solution separated by a semi-permeable membrane from the solvent? |
| Answer» It will result into reverse osmosis, i.e., there will be net flow of the solvent from the solution to the solvent. | |
| 1993. |
The freezing point of `1%` aqueous solution of calcuim nitrate will beA. `0^(@)C`B. Above `0^(@)C`C. `1^(@)C`D. Below `0^(@)C` |
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Answer» Correct Answer - D Freezing point is lowered on addition of solution in it. |
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| 1994. |
A perfectly semi-permeable membrane when used to separate a solution from its solvent permits through it the passage ofA. Solute onlyB. Solvent onlyC. Both (a) and (b)D. None |
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Answer» Correct Answer - B Perfectly semi-permeable membrane allows the passage of solvent particles only. |
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| 1995. |
The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If `x_(A)` and `y_(A)` are the mole fractions of A in the liquid and vapour, respectively find the value of `x_(A)` for which `(y_(A)-x_(A))` has maximum. What is the value of the pressure at this composition? |
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Answer» Since `y_(A) = (x_(A)P_(A)^(@))/(P_(B)^(@)+(P_(A)^(@) -P_(B)^(@))x_(A))` Substracting `x_(A)` from both the sides, we get `y_(A) = x_(A) = (x_(A)P_(A)^(@))/(P_(B)^(@)+(P_(A)^(@)-P_(B)^(@))x_(A)) -x_(A)` Differentiating this with respect to `x_(A)`, we get `(d(y_(A)-x_(A)))/(dx_(A)) =(P_(A)^(@))/(P_(A)^(@)+(P_(A)^(@)-P_(B)^(@))x_(A))- (x_(A)P_(A)^(@)(P_(A)^(@)-P_(B)^(@)))/({P_(B)^(@)+(P_(A)^(@)-P_(B)^(@))x_(A)}^(2)) -1` The value of `x_(A)` at which `y_(A) -x_(A)` has a maximum value can be obtained by setting the above differential equal to zero. Thus, we have `(P_(A)^(@))/(P_(A)^(@)+(P_(A)^(@)-P_(B)^(@))x_(A)) -(x_(A)P_(A)^(@)(P_(A)^(@)-P_(B)^(@)))/({P_(B)^(@)+(P_(A)^(@)-P_(B)^(@))x_(A)}^(2)) -1 =0` Solving for `x_(A)`, we get `x_(A) = (sqrt(P_(A)^(@)P_(B)^(@))-P_(B)^(@))/(P_(A)^(@)-P_(B)^(@))` The value of `P` at this composition is `P = x_(A) P_(A)^(@) +x_(B) P_(B)^(@)` or `P = P_(B)^(@) +(P_(A)^(@) -P_(B)^(@)) x_(A)` or `P = P_(B)^(@) +(P_(A)^(@) -P_(B)^(@)) ((sqrt(P_(A)^(@)P_(B)^(@))-P_(B)^(@))/(P_(A)^(@)-P_(B)^(@)))` or `P = sqrt(P_(A)^(@)P_(B)^(@))` |
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| 1996. |
For an ideal binary solution (A) and (B), `y_(A)` is the mole fraction of A in vapour phase. Which of the following plot should be linear?A. `P_("Total") vs y_(A)`B. `P_("Total") vs y_(B)`C. `(1)/(P_("Total")) vs y_(A)`D. `(1)/(P_("Total"))vs (y_(B)+y_(A))` |
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Answer» Correct Answer - C `P_("Total") = x_(A)p_(A)^(@) +x_(B)p_(B)^(@)` `= x_(A)p_(A)^(@) +(1-x_(A))p_(B)^(@)` `= x_(A) (p_(A)^(@) -p_(B)^(@)) +p_(B)^(@)` `x_(A) = (P_("Total")-p_(B)^(@))/((p_(A)^(@)-p_(B)^(@)))` ...(i) Now, `y_(A) = (p_(A))/(P_("Total")) = (x_(A)p_(A)^(@))/(P_("Total"))` `x_(A) = (y_(A)P_("Total"))/(p_(A)^(@))` ...(ii) Eliminate `x_(A)` between (i) and (ii) `(P_("Total")-p_(B)^(@))/(p_(A)^(@)-p_(B)^(@))=(y_(A)P_("Total"))/(p_(A)^(@))` `(P_("Total")-p_(B)^(@))/(P_("Total"))=((p_(A)^(@)-p_(B)^(@))/(p_(A)^(@)))y_(A)` `1 -(p_(B)^(@))/(P_("Total")) = ((p_(A)^(@)-p_(B)^(@))/(p_(A)^(@)))y_(A)` `(p_(B)^(@))/(P_("Total")) = 1- ((p_(A)^(@)-p_(B)^(@))/(p_(A)^(@)))y_(A)` `(1)/(P_("Total")) =(1)/(p_(B)^(@)) -(p_(A)^(@)-p_(B)^(@))/(p_(A)^(@)p_(B)^(@))` `(1)/(P_("Total"))` is a linear function of `y_(A)`. |
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| 1997. |
Solubility of deliquescent substances in water is generally:A. highB. lowC. moderateD. cannot be predicted |
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Answer» Correct Answer - A A deliquescent substance absorbs large amount of water hence solubility is very high |
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| 1998. |
What happens to blood cells when they are placed in water containing less than 0.99% (mass/volume) salt? |
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Answer» The blood cells collapse due to loss of water by osmosis when placed in water containing less than 0.9% (mass/volume) salt. |
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| 1999. |
Why is molality generally preferred over molarity as unit for expressing concentration of solutions? |
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Answer» Molality is preferred over molarity as unit for expressing concentration of solutions because it involves mass term which is not affected by change in temperature while molarity involves volume term which changes with temperature. |
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| 2000. |
What is the molar concentration of solute particles in human blood if the osmotic pressure is 7.2 atm at normal body temperature, i.e. 37°C? |
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Answer» Л = CRT or C = Л/RT Л = 7.2 atm R = 0.0821 L atm K-1 mol-1 T = 37°C = 37 + 273 = 310 K. Molar concentration (C) = (7.2 atm)/(0.082L Latm K-1mol-1 ) x (310 K) = 0.283 mol-1 = 0.283 M. |
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