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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.
| 1001. |
Why do gases nearly always tend to be less soluble in liquids as the temperature is raised ? |
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Answer» As the temperature of gaseous solution in liquid increase the kinetic energy of molecules escape out from solution and show low solubility at high temperature. |
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| 1002. |
Heptane and octone form ideal solution. At 373 K, the vapour pressure of the two liquid components are 105.2 K Pa and 46.8 K Pa respectively. If the solution contains 25 g of heptane and 25 g of octane, calculate : Vapour pressure exerted by hepthne Vapour pressure exerted by octane Vapour pressure exerted bythe solution Mole fraction octane in the vapour phase. |
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Answer» `"No. of moles of heptane" (n_(B))=("Mass of heptane"(C_(7)H_(16)))/("Gram molar mass")=((25g))/((100 g mol^(-1)))=0.25 mol` `"No. of moles of cotanae" (n_(A))=("Mass of heptane"(C_(8)H_(18)))/("Gram molar mass")=((35g))/((114 g mol^(-1)))=0.307 mol` `"Molar fraction of hepthane" (x_(B))=n_(B)/(n_(B)+n_(A))=((0.25 mol))/((0.25 mol+0.307 mol))=0.449` `"Molar fraction of octane" (x_(A))=n_(A)/(n_(B)+n_(A))=((0.307 mol))/((0.25 mol+0.307 mol))=0.551` `"Vapour pressure of pure heptane" (P_(B)^(@))=105.2 kPa` `"Vapour pressure of pure octane" (P_(A)^(@)) = 46.8 kPa` ` (i) "Vapour pressure of heptane" (P_(B)^(@))=P_(B)^(@)x_(B)=(105.2 kPaxx0.449)47.23 kPa` ` (ii) "Vapour pressure of octane" (P_(A))=P_(A)^(@)x_(A)= (47.23+25.79=73.02 kPa)` `(iii) "Total vapour pressure of solution "(P)=P_(A)+P_(B)=(47.23+25.79)=73.02 kPa` (iv)"mole fraction of octane in the vapour phse"= `P_(A)/(P_(A)+P_(B))=((25.79K Pa))/((73.02 K Pa))= 0.353` |
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| 1003. |
Why do gases always tend to be less soluble in liquids as the temperature is raised? |
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Answer» The dissolution of a gas in a liquid is exothermic in nature because the gas contracts in volume. `"Gas" + "Liquid" hArr "Dissolved gas" DeltaH=-ve` An increase in tmperature will facour the reverse process since it is of endthermic nature. Therefore, the solubility of the gas in solution decreases with the rise in tmeperature. |
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| 1004. |
Assertion:In an ideal solution , `Delta_"mix"H` is zero Reason :In an ideal solution , A-B interactions are lower than A-A and B-B interactions.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - C In an ideal solution, A-B interactions are same as A-A and B -B interactions. |
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| 1005. |
Assertion:The vapour pressure of an aqueous solution of sucrose is less than 1.013 bar at 373.15 K Reason : Vapour pressure of water is 1.013 bar at 373.15 K.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - B Vapour pressure of solution containing non-volatile solute is less than that of pure solvent. |
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| 1006. |
An example of colligative property isA. Boiling pointB. OsmosisC. Freezing pointD. Osmotic pressure. |
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Answer» Correct Answer - D Osmotic pressure depends upon the no of particles in solution. |
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| 1007. |
Assertion:Lowering of vapour pressure is not dependent on the number of species present in the solution. Reason : Lowering of vapour pressure and relative lowering of vapour pressure are colligative propertiesA. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - D Lowering of vapour pressure is directly proportional to the number of species present in the solution. Only relative lowering of vapour pressure is a colligative property |
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| 1008. |
Which of the following is not a colligative property?A. `DeltaT_(f)`B. `pi`C. `DeltaT_(b)`D. `K_(b)` |
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Answer» Correct Answer - D `k_(b)` is molal elevation constant. |
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| 1009. |
The solution A is twice hypertonic to the solution B at a given temperature. If the solution A contains 8.6 × 1022 molecules, then the number of molecules present in B are,(a) 8.6 × 1022 (b) 1.73 × 1022 (c) 3.24 × 1022 (d) 4.3 × 1022 |
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Answer» Option : (d) 4.3 × 1022 |
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| 1010. |
200mL of an aqueus solution of a protien 1.26 g of the protein. The osmotic pressure of such a solution is found to be 2.7xx10^(-3) bar at 300 K. Calculate the molar mass of protein `(R=0.083 L bnar mol^(-1)K^(-1))` |
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Answer» Osmtic pressure `(pi)=(W_(B)xxRxxT)/(M_(B)xxV) or M_(B)=(W_(B)xxRxxT)/(pixxV)` `pi=2.7xx10^(-3) bar, V=200mL=200/1000=0.2L.` `W_(B)=1.26g, R=0.083 L bar mol^(-1)K^(-1), T=300K` `M_(B)=((1.26g)xx(0.083 L bar mol^(-1)K^(-1))xx(300K))/((2.7xx10^(-3)bar)xx(0.2L))=58,100g mol^(-1)` |
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| 1011. |
0.1m solution of KCI and `BaCI_(2)` are preparred. The freezing point of KCI solution is found to be- `4.0^(@)C`. What will be the freezing point of `BaCI_(2)` solution assuming that both KCI and `BaCI_(2)` solution are sompletely ionised in solution ?A. `-3^(@)C`B. `-4^(@)C`C. `-5^(@)C`D. `-6^(@)C` |
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Answer» Correct Answer - a `"For" KCI,DeltaT_(f)=iK_(f)xxm` `DeltaT_(f)=(0.0.4^(@)C)=4^(@)C` `4K=2xxK_(f)xxm` `"For " BaCI_(2),DeltaT_(f)=iK_(f)xxm` `=3xxK_(f)xxm` Both `K_(f)` and m are same for the solute, `DeltaT_(f)=(BaCI_(2))=3` Freezing point of `BaCI_(2)=0-3=-3^(@)C` |
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| 1012. |
Which of the following is dependent of temperature ?A. MolalityB. MolarityC. Mole fractionD. Mass precent. |
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Answer» Correct Answer - b Molarity (M) of a solution is tempreture dependent since it takes into consideretion the volume of the solution. |
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| 1013. |
An aqueous solution of glucose containing 60 g glucose `(C_(6)H_(12)O_(6))` per litre has an osmotic pressure of 5.2 bar at 300 K. The concentration of the glucose solution having osmotic pressure 0of 1.3 bar at the same temperature is :A. `1/10M`B. `1/5M`C. `/220M`D. `1/12M` |
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Answer» Correct Answer - e Osmotic pressure `(pi)alphaM` `pi_(1)/pi_(2)=M_(1)/M_(2)` (for the same solution at the same temperature) `(5.2)/(1.3)=(1/3)/M_(2)or M_(2)=1/3xx1/4=1/12M` |
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| 1014. |
The molal depression constant depends uponA. Nature of soluteB. Nature of solventC. `DeltaH_("solution")`D. Vapour pressure of solution |
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Answer» Correct Answer - B `k_(b)` depends on the nature of solvent, its mol mass and b.p |
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| 1015. |
One liter of aqueus solution of suctrose `("molar mass"=324 g mol^(-1))` Weighing 1015 g is found g is foung to record an osmotic pressure of 4.82 atm at 293 K. What is the molality of the solution ? |
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Answer» Calculation of number of moles of solute. `pi(nRT)/V or n=(piV)/(RT)` `pi=4.82 atm, V=1L, R=0.0821 L atm K^(-1) mol^(-1),T=293 K` `n=((4.82 atm)xx(1L))/((0.0821 L atm K^(-1)mol^(-1))xx(293K))=0.2 mol.` Calculation of the molality of solution. Mass of solute (sucrose)=`(0.2mol)xx(342 g mol^(-1))=68.4 g` Mass of solution=1015 g Mass of solvent(water)=1015-68.4=946.6 g =0.9466 kg Molality of solution (m)= `(Mass of glucose//Gram molar mass)/(Mass of solvent in kg) =((68.4g))/((342g mol^(-1)xx(0.9466 kg))=0.2113 mol//kg)= 0.2113 m` |
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| 1016. |
Osmotic pressure of a solution at 300 K is 73.8 atm. Assuming non-electrolytic solute, the molarity of solution will be : |
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Answer» Correct Answer - 3 `pi = CRT` `73.8 = C xx 0.082 xx 300` `C = 3 M`] |
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| 1017. |
The osmotic pressure of a solution containing `5 g` of substance (molar mass =100) in 308 m L of solution was found to be 4.0 atm at `300 K`. Calculate the value of solution constant (R) |
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Answer» The given values are `W_(B)=5 g`,`Mw_(B)=100 mL^(-1)` `pi=4.0 atm`,`T=300 K`,`V=380 mL` Using formula, `R=piV/piT` `=4.0xx(308/1000)/(5/100)xx300=0.0821 L-"atm" "mol"^(-1) "degree"^(-1)` |
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| 1018. |
The osmotic pressure of a solution was found to be `8 atm` when `8 mol ` of a non-volatile solute was dissolved in `V` L of solution at `300 K`.Calculate the volume of solution `(R=0.0821 L-"atm" K^(-1) "mol"^(-1))` |
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Answer» The given values are `pi=8 atm`,`T=300K`,`n_(B)=8` `R=0.0821 L-"atm mol"^(-1)` Now,using formula `V=(nRT)/(pi)=(8xx0.0821xx300)/8 =246.3 L` |
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| 1019. |
A solution of cane sugar containing 18 g L-1 has an osmotic pressure 1.25 atm. Calculate the temperature of the solution.(Molar mass of cane sugar = 342, R = 0.082 lit atm mol-1K-1) |
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Answer» Given : Amount of cane sugar = W = 18 g L-1 Osmotic pressure = π = 1.25 atm Molar mass of cane sugar = M = 342 g mol-1 Temperature = T = ? Number of moles of cane sugar = \(\frac{W}{M}\) = \(\frac{18}{342}\) = 0.05263 mol lit-1 ∴ Concentration of solution = C = \(\frac{n}{V}\) = \(\frac{0.05263}{1}\) = 0.05263 mol lit-1 π = CRT ∴ T = \(\frac{\pi}{CR}\) = \(\frac{1.25}{0.05236\times 0.08206}\) = 289.4 K ∴ Temperature of solution = 289.4 K |
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| 1020. |
Calculate the osmotic pressure of 0.2 M glucose solution at 300 K. (R = 8.314 J mol K-1) |
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Answer» Given : Concentration of the solution = C = 0.2 M = 0.2 mol dm-3 = 0.2 × 10 mol m-3 Temperature = T= 300 K = 8.314 J mol-1 K-1 The osmotic pressure, π is given by, π = CRT = 0.2 × 103 × 8.314 × 300 = 4.988 × 105 Nm-2 (or Pa) ∴ Osmotic Pressure = 4.988 × 105Nm-2 |
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| 1021. |
A solution of 2.5g of non-volatile solid in 100g benzene is boiled at `0.42^(@)C` higher than the boiling point of pure benzene. Calculate the molecular mass of the substance. Molal elevation constant of benzene is `2.67 "K kg mol"^(-1)`. |
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Answer» `m=(1000K_(b)xxw)/(WxxDeltaT)` Given, `K_(b)=2.67,w=2.5g,W=100g,DeltaT=0.42` `m=(1000xx2.67xx2.5)/(100xx0.42)=158.9` The molecular mass of substance is 158.9. |
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| 1022. |
An aqueous solution of glucose is made by dissolving 10 g of glucose in 90 g water at 303 K. If the V.P. of pure water at 303 K be 32.8 mm Hg, what would be V.P. fo the solution ? |
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Answer» `(32.8-P_(s))/(32.8)=((10)/(60))/((10)/(60)+(90)/(18))` `(32.8-P_(s))/(32.8)=(0.16)/(0.16+5)` `(32.8-P_(s))/(32.8)=0.03` `32.8-P_(s)=1.017` `P_(s)=31.7" atm"` |
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| 1023. |
represents the distillation of mixture of liquid `A` and liquid `B` which gives both of pure liquid `A` and `B` . Represents the azeotropic mixture of `HNO_(3)` and `H_(2)O` which distillation gives an azeotropic mixture and either of pure liquid. We cannot separate both the pure liquid, i.e., `H_(2)O` and `HNO_(3)`. At temperature `T_(1)` and composition `Q`, which of the following is true? A. Vapour phase is richer in `B` while liquid phase is richer in `A`.B. Distillation of composition `Q` gives only pure `A`.C. Distillation of composition `Q` gives only pure `A` and pure `B`.D. Distillation of composition `Q` gives higher percentage of `B` and `A`. |
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Answer» Correct Answer - A::C::D Composition `Q` lies closer to component `B`, and at temperature `T_(1)`, vapour phase will contain more of `B`. |
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| 1024. |
represents the distillation of mixture of liquid `A` and liquid `B` which gives both of pure liquid `A` and `B` . Represents the azeotropic mixture of `HNO_(3)` and `H_(2)O` which distillation gives an azeotropic mixture and either of pure liquid. We cannot separate both the pure liquid, i.e., `H_(2)O` and `HNO_(3)`. What is the result of distilling a mixture of `80% HNO_(3)` and `20%H_(2)O`? A. Pure `H_(2)O` and azeotropic mixture can be separated.B. Pure `H_(2)O` and pure `HNO_(3)` can be separated.C. Pure `HNO_(3)` and azeotropic mixture can be separated.D. None of these |
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Answer» Correct Answer - C The composition of `80% HNO_(3)` and `20% H_(2)O` lies in between the azeotropic mixture and pure `HNO_(3)`. Therefore , distillation of such composition will give pure `HNO_(3)` and azeotropic mixture. |
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| 1025. |
Which relations are not correct for an aqueous dilute solution of `K_(3)PO_(4)` if its degree of dissociation is `alpha`?A. `DeltaP/P^(@)=(Molalityxx18xx(1+3alpha))/1000`B. `DeltaP/P^(@)=(pi_(obs)xx18xx(1+3alpha))/(RT xx 1000)`C. `DeltaP/P^(@)=(DeltaT_(f)_(obs)xx18)/(K_(f) xx 1000)`D. Mw of `K_(3)PO_(4))=Mw_(obs)xx(1+3alpha)` |
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Answer» Correct Answer - A::C::D `DeltaP/P^(@)=n_(2)/n_(1)=(n_(2) xx Mw_(1) xx 1000)/(W_(1) xx 1000)=("Molality" xx Mw_(1))/1000` For electrolyte `(DeltaP)/P^(@)=(Molality xx M)/(1000) xx (1 + 3alpha)` `(Mw_(1)=18 "for" H_(2)O)` also, `pi_(obs)=C xx R xxT (1 + 3alpha)` `:.DeltaP/P^(@)=(pi_(obs))/(RT) xx 18/1000` `DeltaT_(f obs)=K_(f) xx "molality" xx (1+ 3alpha)` `(DeltaP)/P^(@)=(DeltaT_(f obs) xx 18)/(K_(f) xx 100)` `i=(1 + 3alpha)="Calculated molecular weight"/"Observed molecular weight"` Therefore, molecular weight of `K_(3)PO_(4)=M_(obs) xx (1 + 3alpha)` |
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| 1026. |
represents the distillation of mixture of liquid `A` and liquid `B` which gives both of pure liquid `A` and `B` . Represents the azeotropic mixture of `HNO_(3)` and `H_(2)O` which distillation gives an azeotropic mixture and either of pure liquid. We cannot separate both the pure liquid, i.e., `H_(2)O` and `HNO_(3)`. What is the result of distilling a mixture of `50% HNO_(3)` and `50%H(2)O` ? A. Pure water and azeotropic mixtue can be separated.B. Pure `H_(2)O` and pure `HNO_(3)` can be separated.C. Pure `HNO_(3)` and azeotropic mixture can be separated.D. None of these |
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Answer» Correct Answer - A The composition of `50% H_(2)O` and `50% HNO_(3)` will lies in between the azeotropic mixture and pure `H_(2)O`. Therefore , distillation of such composition will give pure `H_(2)O` and azeotropic mixture. |
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| 1027. |
For associative solutesA. `I lt 1` and `alpha lt 1`B. `i gt 1` and `alpha gt 1`C. `I lt 1` and alpha gt 1`D. `I gt 1` and `alpha lt 1` |
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Answer» Correct Answer - 1 |
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| 1028. |
The liquid used in car radiator is primarity a mixture of ethylene glycol is thatA. It helps in smooth combinationB. It lowers the boiling point of waterC. It causes the freezing point to decreaseD. It is more volatile |
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Answer» Correct Answer - 3 |
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| 1029. |
Sodium chloride solution freezes at lower temperature then water but boils at higher temperature than water. Explain. |
| Answer» The freezing point of a liquid depresses on the addition of a non-volatile solute, and therefore, a solution of sodium chloride freezes at a temperature lower than the freezing point of water. On the other hand, there is elevation of boiling point on the addition of a non-volatile solute and consequently the boiling point of sodium chloride solution is more then that of water. | |
| 1030. |
Calculate the boiling point of a solution of `18.2 g` DDT `(C_(14)H_(9)Cl_(5))`, a nonvolatile, non electrolytes substance, in 342 g of chloroform, `CHCl_(3). K_(b)` for `CHCl_(3)` is `3.63 Kkgmol^(-1)` and boiling point `(bp)` is `334.9 K`. Strategy: First find the increase in boiling point form the relationship. `DeltaT_(b)=K_(b)m`. The boiling point of solution is bigger by this amount than the normal boiling point of pure `CHCl_(3)`. |
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Answer» Calculate the molality of DDT in the solution The molecular mass of DDT is `354.5 u`. Therefore `n_(DDT)=(mass_(DDT))/(molar mass_(DDT))` `=(18.2 g)/(354.5 g mol^(-1))=0.0513 mol` The molality is given by `m=n_(DDT)/g_(solv)xx(1000 g)/(kg)` `=(0.051 mol DDT)/(342 g CHCl_(3))xx(1000 g)/(kg)` `=0.149 mol Kg^(-1)` Step 2. Calculate the boiling point elevation of chloroform that is caused by the addition of `18.2 g DDT`. Use the molality calculated here and the value of `K_(b)` for `CHCl_(3)` into equation `(2.64)` `DeltaT_(b)=K_(b)m` `=(3.63 KKg mol^(-1))(0.149 mol Kg^(-1))` `=0.54 K` Step 3. Calculate the boiling point for the `CHCl_(3)` in the solution. The boiling point of pure `CHCl_(3)` is raised by `DeltaT_(b)` in this solution. `DeltaT_(b)=T_(b)-T_(b)^(0)` or `T_(b)=T_(b)^(0)+DeltaT_(b)` `=(334.9 K)+(0.54 K)` `=335.4 K` |
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| 1031. |
Why is it advised to add ethylene glycol to water in car radiator while driving in a hill station? |
| Answer» Ethylene glycol lowers the freezing point of water, and therefore, it does not freeze in a hill station. | |
| 1032. |
A solution containg `12 g` of a non-electrolyte substance in `52 g` of water gave boiling point elevation of `0.40 K` . Calculate the molar mass of the substance. `(K_(b)` for water =` 0.52 K kg mol^(-1))` |
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Answer» The given values are: `W_(B) = 12 g`,`W_(A) = 52 g` `DeltaT_(b) = 0.4 K` Using the formula, `Mw_(B) = (K_(b) xx W_(B) xx 1000) / (DeltaT_(b) xx W_(A))` `:. Mw_(B) = (0.52 xx 12xx 1000) /(0.4 xx 52 ) = 300` Molecular weight of solute is `300 g mol^(-1)`. |
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| 1033. |
A solution of 12.5 g of an unknown solute in 170 g water gave a boiling point elevation of 0.63 K. Calculate the molar mass of the solute. `(K_(b)=0.53 K kg mol^(-1))`. |
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Answer» Correct Answer - 61.85 g `mol^(-1)` `W_(B)=12.5g, W_(A)=0.170 kg, DeltaT_(b)=0.63 K, K_(b)=0.53" K kg mol"^(-1), M_(B)=?` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.53" K kg mol"^(-1))(12.5g))/((0.63 K)xx(0.170 kg))=61.85" g mol"^(-1)`. |
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| 1034. |
Ebullioscopic constant `K_(b)` corresponds to the change in boiling point produced by a one ___ ideal solution of a nonvolatile nonelectrilyte.A. molarB. molalC. normalD. formal |
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Answer» Correct Answer - 2 The elevation of nonvolatile point produced by dissolving `1 mol` of any nonvolatile solute, which does not dissociate or associate, in `1 kg` of a particular solvent is known as the ebillioscopic constant for that solvent. It is the ratio of the elevation in `BP` to molality, i.e., `K_(b)=DeltaT_(b)//m` Thus, the units of `K_(b)` are `.^(@)C//m` or `K//m` or `KKg mol^(-1)`. |
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| 1035. |
One molal solution of a given solvent is always less concentrated than one molar solution. Explain. |
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Answer» In one molar solution one gram mole of solute is dissolved in one litre of solution while in case of one molal solution same one gram mole of solute is dissolved in 1000 gm of solvent only, which on considering normal density parameters of water, can’t be lesser in amount than solvent part present in one litre solution. Therefore,more amount of solvent is present in one molal solution than in one molar solution. |
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| 1036. |
Match the terms given in Column I with expressions given in Column II.Column IColumn II(i) Mass percentage (a) Number of moles of the solute component/Volume of solution in litres(ii) Volume percentage(b) Number of moles of a component/Total number of moles of all the components(iii) Mole fraction(c) Volume of the solute component in solution/Total volume of solution x 100(iv) Molality(d) Mass of the solute component in solution/Total mass of the solution x 100 (v) Molarity(e) Number of moles of the solute components/Mass of solvent in kilograms |
| Answer» (i) → (d) (ii) → (c) (iii) → (b) (iv) → (e) (v) → (a) | |
| 1037. |
Assertion (A) Molarity of a solution in liquid state changes with temperature. Reason (R ) The volume of a solution charges with change in temperature.A. Assertion and reason both are correct statements and reason is correcft explanation for assertion.B. Assertion and reason both are correct statements but reason is not correct explanation for assertion.C. Assertion is correct statement but reason is wrong statement.D. Assertion and reason both are incorrect statements. |
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Answer» Correct Answer - a Assertion and reason both are correct statements and reason is the correct explanation of assertion. Volume of solutions is a function of temperature which varise with temperatur. Hencemolarity of solution in liquid state changes with temperature. `" "Molarity =("moles of solute")/("volume of solution in litre")` |
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| 1038. |
Can azeotropes be separated by fractional distillation? |
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Answer» No, azeotropes can’t be separated by fractional distillation. |
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| 1039. |
Define ebullioscopic constant and give its units. |
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Answer» Ebullioscopic constant is defined as the elevation in boiling point of a solution of a non-volatile solute when its molality is unity. Its units are K Kg mol-1. |
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| 1040. |
When a non-volatile solute is added to solvent,there is increase in boiling point of solution. Explain. |
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Answer» When a non-volatile solute is added to a volatile solvent the vapour pressure of pure solvent decreases because a part of the surface is occupied by non-volatile solute which can’t volatilise. As a result, the vapour pressure of solution decreases and hence, the solution requires a comparatively higher temperature to boil causing an elevation of boiling point. |
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| 1041. |
In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.(i) Assertion and reason both are correct statements and reason is correct explanation for assertion.(ii) Assertion and reason both are correct statements but reason is not correct explanation for assertion.(iii) Assertion is correct statement but reason is wrong statement.(iv) Assertion and reason both are incorrect statements.(v) Assertion is wrong statement but reason is correct statement.Assertion : Molarity of a solution in liquid state changes with temperature.Reason : The volume of a solution changes with change in temperature |
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Answer» (i) Assertion and reason both are correct statements and reason is correct explanation for assertion. |
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| 1042. |
The diagram given below represents boiling point composition diagram of solution of component `A` and `B`, which is/are incorrect among the following? A. The solution shows negative deviationB. `A-B-` interactions are stronger than `A-A` and `B-B`C. `K_(H)` has temperature dependenceD. `K_(H)` increases with temperature |
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Answer» Correct Answer - A::B::C For`-ve` deviation `{:(A--B gt A--A),(A--B gt B--B):}` `-ve` diviation solution are non ideal solution. |
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| 1043. |
Assertion (A) When methyl alcohol is added to water, boiling point of water increases. Reason (R ) When a volatile solute is added to a volatile solvent evevation in boiling point is observed.A. Assertion and reason both are correct statements and reason is correcft explanation for assertion.B. Assertion and reason both are correct statements but reason is not correct explanation for assertion.C. Assertion is correct statement but reason is wrong statement.D. Assertion and reason both are incorrect statements. |
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Answer» Correct Answer - d Assertion is wrong statement but reason is correct statemet When methyl alcohol is added to water, boiling point of water decreases because when a volatile solute is added to a volatile solvent elevation in boiling point is observed. |
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| 1044. |
In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.(i) Assertion and reason both are correct statements and reason is correct explanation for assertion.(ii) Assertion and reason both are correct statements but reason is not correct explanation for assertion.(iii) Assertion is correct statement but reason is wrong statement.(iv) Assertion and reason both are incorrect statements.(v) Assertion is wrong statement but reason is correct statement.Assertion : When methyl alcohol is added to water, boiling point of water increases.Reason : When a volatile solute is added to a volatile solvent elevation in boiling point is observed. |
| Answer» (iv) Assertion and reason both are incorrect statements. | |
| 1045. |
What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm3 of water? Why? |
Answer»
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| 1046. |
Which of the four colligative properties is most often used for molecular mass determination? Why? |
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Answer» 1. Since osmotic pressure has large values, it can be measured more precisely. 2. The osmotic pressure can be measured at a suitable constant temperature. 3. The molecular masses can be measured more accurately. 4. Therefore, it is more useful to determine molecular masses of expensive substances by osmotic pressure. |
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| 1047. |
Relative vapour pressure lowering depends only on : (a) Mole fraction of solute (b) Nature of solvent (c) Nature of solute (d) Nature of solute and solvent |
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Answer» Option : (a) Mole fraction of solute |
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| 1048. |
A solution having mole fraction of a solute equal to 0.05 has vapour pressure 20 × 103 Nm-2. Hence the vapour pressure of a pure solvent is :(a) 21.05 × 103 Nm-2 (b) 4 × 105 Nm-2 (c) 1 × 103 Nm-2 (d) 2 × 104 Nm-2 |
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Answer» Option : (a) 21.05 × 103 Nm-2 |
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| 1049. |
The relative lowering of vapour pressure of a solution is proportional to the :(a) mole fraction of the solvent (b) mole fraction of the solute (c) amount of the substance (d) volume of the solvent |
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Answer» Option : (b) mole fraction of the solute. |
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| 1050. |
Raoult’s law is not applicable to : (a) solutions of volatile solutes (b) solutions of electrolytes (c) dilute solutions (d) concentrated solutions |
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Answer» Option : (d) concentrated solutions |
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