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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.
| 1151. |
Which of the following is less than zero for ideal solutions ?A. `DeltaH_("mix")`B. `DeltaV_("mix")`C. `DeltaG_("mix")`D. `DeltaS_("mix")` |
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Answer» Correct Answer - C For an ideal solution ,`DeltaG "mix" lt 0` |
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| 1152. |
A maximum or minima obtained in the temperature, composition curve of a mixture of two liquids indicates:A. an azeotropic mixtureB. an eutectic formationC. that the liquids are immiscible with one anotherD. that the liquids are partially miscible at the maximum or minimum |
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Answer» Correct Answer - A An azeotropic mixture boil at particular temperature without changing its composition. |
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| 1153. |
A `5.00 g` sample of vinegar is titrated with `0.108 M NaOH`. If the vinger requires `39.1 mL` of the `NaOH` for complete reaction, the mass percentage of acetic acid `(CH_(3)CO_(2)H)` in the vinegar isA. 0.0605B. 0.0407C. 0.0786D. 0.0506 |
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Answer» Correct Answer - 4 The reaction is `CH_(3)CO_(2)H(aq.)+NaOH(aq.) rarr CH_(3)CObar(O)Na^(+)(aq.)+H_(2)O(l)` Convert the volume of NaOH to moles of NaOH using the molarity formula. Then we convert moles NaOH to moles `CH_(3)CO_(2)H` using the balanced chemical equation. Finally we covert moles `CH_(3)CO_(2)H` to grams `CH_(3)CO_(2)H`. Mass percentage of acetic acid in the vinegar `=("Mass of acetic acid")/("Mass of vinegar")xx100` `n_(NaOH)=V_(NaOH "soln")xxM_(NaOH "soln")` `=(39.1xx10^(-3)L)(0.108 mol L^(-1))` `=4.22xx10^(-3) mol` According to equation, moles of `CH_(3)CO_(2)H` and `NaOH` reacting are equal. Thus `n_(CH_(3)COOH)=n_(NaOH)=4.22xx10^(-3) mol` `mass_(CH_(3)CO_(2)H)=n_(CH_(3)CO_(2)H)xx "molar mass"_(CH(3)CO_(2)H)` `=(4.22xx10^(-3) mol)(60 g mol^(-1))` `=0.253 g` Thus mass% `=(0.253 g)/(5.00 g)xx100%=5.06%` |
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| 1154. |
An organic compound `C_xH_(2y)O_y` was burnt with twice the amount of oxygen needed for complete combustion to `CO_2` and `H_2O`. The hot gases when cooled to `0^@C` and 1 atm pressure, measured `2.24` litre. The water collected during cooling weighed 0.9 gm. the vapour pressure of pure water at `20^@C` is 17.5 mm Hg and is lowered by 0.104 mm when 50 gm of the organic compound is dissolved in 1000 gm of water. Give the molecular formula of the organic compound. |
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Answer» Correct Answer - `C_(5)H_(10)O_(5)` `C_xH_(2y)O_y+2xO_2 hArrxCO_2+yH_2O+xO_2` After cooling only `CO_(2)` and `O_(2)` are present because water is in liquid form. Thus, moles of gases after cooling =`x + x =2x` Volume of gases after cooling =`2.24 L` `2x=2.24 L` `x=1.12 L` Number of moles of `CO_(2)=1.12/2.24 =0.5 mol` Hence, the empirical formula of organic compound is `C(H_(2)O)` Given, Vapour pressure of pure water `(P^(@))=17.5 mm Hg` Lowering of vapour pressure `(P^(@)-P)=0.104 mm Hg` weight of solute `(W_(2)) =50 g` weight of solvent `(W_(1)) =1000 g` `(P^(@)-P)/ P^(@)=(W_(2) xx Mw_(1))/(Mw_(1) xx W_(1))` `0.104/17.5=(50 xx 18)/(Mw_(2) xx 1000)` `Mw_(2)=(900 xx 17.5)/104=151.4 g` Empirical formula weight =`C(H_(2)O)` =`12 +2 +16 =30` Molecular weight (m)=151.4 g` `n=("Molecular weight")/("Empirical formula weight")` =`151.4/30 =5.04 =5` Thus, molecular formula =`[C(H_(2)O]_(5)=C_(5)H_(10)O_(5)` |
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| 1155. |
Which one of the following is applied in water purification? |
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Answer» reverse osmosis |
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| 1156. |
In commercial reverse osmosis process, the semi permeable membrane used is …. |
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Answer» In commercial reverse osmosis process, the semi permeable membrane used is cellulose acetate. |
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| 1157. |
Give an example of Ideal solution. |
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Answer» n-Hexane and n-Heptane or any other suitable example. |
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| 1158. |
How would you define, an Ideal solution? |
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Answer» The solution which obey Raoult’s law at all temperature and pressure is called an ideal solution which also have ∆mixH and ∆mixV = 0 |
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| 1159. |
What is molarity? |
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Answer» No. of moles of solute dissolved per litre of a solution |
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| 1160. |
State Henry’s law. |
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Answer» The partial pressure of a gas is directly proportional to its mole fraction in solution. P = KH x X |
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| 1161. |
What do you mean by vapour pressure? |
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Answer» The pressure exerted by vapours above the liquid surface at equilibrium is called vapour pressure. |
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| 1162. |
The volume of water to be added to `N//2 HCI` to prepare `500 cm^(3)` of `N//10` solution isA. `45cm^(3)`B. `400 cm^(3)`C. `450cm^(3)`D. `100cm^(3)` |
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Answer» Correct Answer - B `N_(1)V_(1) = N_(2)V_(2)` `(N)/(2) xx V_(1) = (N)/(10) xx 500` `V_(1) = 100` Vol of `N//10` solution `= 500 cm^(3)` Vol. of `N//2` solution `= 100 cm^(3)` `:.` Water to be added `= (500 - 100) cm^(3) = 400 cm^(3)` |
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| 1163. |
How much `K_(2)Cr_(2)O_(7) (M.W. = 294.19)` is required to prepare one litre of `0.1N` solution?A. `9.8063g`B. `7.3548g`C. `3.6774g`D. `4.903g` |
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Answer» Correct Answer - D `Cr_(2)O_(7)^(2-) + 14H^(+) +6e^(-) rarr 2Cr^(3+) +7H_(2)O` Eq. mass of `K_(2)Cr_(2)O_(7) = ("Mol. mass")/(6) = (294.19)/(6)` `=49.03` Mass of `K_(2)Cr_(2)O_(7)` required to prepare 1L of `0.1N K_(2)Cr_(2)O_(7)` solution `= 4.903g` |
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| 1164. |
The ionic strength of a solution containing `0.1"mole"//kg` of KCI and `0.2"mole"//kg` of `CuSO_(4)` isA. 0.3B. 0.6C. 0.9D. 0.2 |
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Answer» Correct Answer - C Ionic strength is half of the sum of all the terms obtained by multiplying the molality of each ion by the square of its valency. `1 = (1)/(2) (m_(1)z_(1)^(2) +m_(2)z_(2)^(2)+m_(3)z_(3)^(2)+m_(4)z_(4)^(2))` Here for `K^(+)` ion, `m_(1) = 0.1 m, z_(1) = 1` For `CI^(-)` ion, `m_(2) = 0.1m, z_(2) = 1` For `Cu^(2+)` ion, `m_(3) = 0.2m, z_(3) = 2` For `SO_(4)^(2-)` ion `m_(4) = 0.2 m, z_(4) = 2` `:. 1 =(1)/(2)[0.1 xx 1^(2) +0.1 xx 1^(2) +0.2 xx 2^(2) +0.2 xx 2^(2)]` `= (1)/(2) [0.1 +0.1 +0.8 +0.8]` `= (1.8)/(2) = 0.9` |
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| 1165. |
Assertion: If solutions of sugar and sodium chloride are prepared by dissolving one mole of each of the solute in 1000g of water they will show different lowering in freezing point. Reason: The colligative properties depend on the nature of the solute.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - C | |
| 1166. |
3.65grams of HCI dissolved in 16.2g of water. The mole fraction of HCI in the resulting solution isA. 0.4B. 0.3C. 0.2D. 0.1 |
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Answer» Correct Answer - D `n(HCI) = (3.65)/(36.5) = 0.1mol` `n(H_(2)O) = (16.2)/(18) = 0.9` mol `X (HCI) = (n(HCI))/(n(HCI)+n(H_(2)O))` `= (0.1)/(0.1+0.9) = 0.1` |
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| 1167. |
What is the relation between normality and molarity of a given solution. |
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Answer» Normality = 2 x molarity. |
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| 1168. |
Write Henry’s law. |
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Answer» The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas at a given temperature. |
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| 1169. |
The boiling point of a solution made by dissolving `12.0 g` of glucose in `100 g` of water is `100.34^(@)C`. Calculate the molecular weight of glucose, `K_(b)` for water `= 0.52^(@)C//m`. |
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Answer» Using the relation for the molecular weight of a solute from elevation in boiling point, we have `Mw_(b) = K_(b)(W_(B)/(W_(A)DeltaT_(b)) xx 1000)` `= 0.52 (12/(100 xx 0.34) xx 1000)` `(DeltaT_(b) = 100.34 - 100 = 0.34^(@)C)` `rArr Mw_(B) = 183.5 g mol^(-1)` |
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| 1170. |
Two liquids A and B boil at 145°C and 190°C respectively. Which of them has a higher vapour pressure of 80°C? |
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Answer» A being more volatile will have higher vapour pressure at 80°C. |
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| 1171. |
What do you understand by colligative properties? |
| Answer» Colligative properties are those properties which depend upon the number of particles of solute but not on the nature of solute. | |
| 1172. |
Two liquids A and B boil at 120°c and 160°c respectively. Which of them has higher vapour pressure at 70° c? |
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Answer» Lower the boiling point, more volatile it is .So liquid A will have higher vapour pressure at 70°c. |
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| 1173. |
Two liquids `A` and `B` boil at `130^(@)C` and `160^(@)C` , respectively. Which of the them has higher vapour pressure at `80^(@)C`. |
| Answer» Liquid `A` with lower boiling point is more volatle and therefore will have higher vapour pressure. | |
| 1174. |
Two liquids `X` and `Y` boil at `110^(@)C` and `130^(@)C`, respectively. Which one of them has higher vapour pressure at `50^(@)C`? |
| Answer» Lower the boiling point, more volatile it is. Hence, liquid `X` will have higher vapour pressure at `50^(@)C`. | |
| 1175. |
Does benzoic acid also exist as dimer in water ? |
| Answer» No, it dissociates into ions in water. It can exist as idmer only in non-polar solvents such as benzene. | |
| 1176. |
`F_(3)CI_(3)` on reaction with `K_(4)[Fe(CN)_(6)]` in aqueous solution gives blue colour. These are separeted by a semipermeable membrene. Will tere be the appearance of a blue colour on the side X due to osmosis ? |
| Answer» Osmosis will take place from te side Y to the side X because 0.01 M `FeCI_(3)` solution is less concentrated. But there will be no formation of blue colour on the side X. Actually, only molecules/particles of water (solvent) will pass through the semipermeable membreane. Neither the `Fe^(3+)` ions nor the `CI^(-)` ions will be able to migrate. Therefore, no chemical reaction will occur and no colour change will take place. | |
| 1177. |
At the same temperature, hydrogen is more soluble in water tham helium. Which of the two has higher `K_(H)` value ? |
| Answer» Helium (He) has higher `K_(H)` value. In genral, more the `K_(H)` value, less is the solubility of a gas in waster. | |
| 1178. |
Two liquids A and B boil `145^(@)C`and `190^(@)C`respectivly. Which of them has a higher vapour pressre at `80^(@)C` ? |
| Answer» Liquid A with lesser boiling point is more volatile as compared to the liquid B and higher vapour presure at `80^(@)C`. | |
| 1179. |
The diameter of colloidal particle is of the orderA. `10^(-3) m`B. `10^(-6) m`C. `10^(-15) m`D. `10^(-7) m` |
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Answer» Correct Answer - B::D Collloidal particle diameter is `10^(-9) m` to `10^(-6) m` |
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| 1180. |
Rate of physisorption increases with :A. decrease in temperatureB. increase in temperatureC. decrease in pressureD. decrease in surface area |
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Answer» Correct Answer - A Physical adsorption is exothermic process so its rate decreases with increase in temperature. |
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| 1181. |
Which of the following colloidal solutions contain negatively charged colloidal particles?A. `Fe(OH)_(3)` solB. `As_(2)S_(3)` solC. BloodD. Gold sol |
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Answer» Correct Answer - B::C::D `Fe(OH)_(3)` is positive sol, remaining all three are negative sol. |
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| 1182. |
Which of the following statements are true for physisorption ?A. Extent of adsorption increases with increase in pressureB. It needs activation energyC. It can be reversed easilyD. It occurs at high temperature |
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Answer» Correct Answer - A::C Physisorption is reversible and its extent increases with pressure. |
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| 1183. |
Which of the following are multimolecular colloids?A. SulphurB. Egg albumin in water (C) Gold solC. Gold solD. Soap solution |
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Answer» Correct Answer - A::C Egg albumin is marcomolecular colloid and soap solution is associated colloid. |
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| 1184. |
Which of the following are incorrect statements?A. Hardy schulz rule is related to coagulationB. Brownian movement and Tyndall effect are the characteristic of colloidsC. In gel, the liquid is dispersed in liquidD. Higher the gold number, more is the protective power of lyophillic sols. |
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Answer» Correct Answer - C::D Gel is liquid in solid dispersion. |
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| 1185. |
Which of the following are hydrophobic sols?A. Protein solB. Gold solC. Gum solD. `Fe(OH)_(3)` sol |
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Answer» Correct Answer - B::D Gold sol and `Fe(OH)_(3)` sol are hydrophobic. |
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| 1186. |
When negatively charged colloids like `As_(2)S_(3)` sol is added to positively charged `Fe(OH)_(3)` sol in suitable amounts :A. Both the sols are precipitated simultaneouslyB. This process is called mutual coagulationC. They becomes positively charged colloidsD. They become negatively charged colloids |
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Answer» Correct Answer - A::B Positive and negative sol will precipitate each other. |
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| 1187. |
Which of the following are examples of aerosols?A. Whipped creamB. CloudC. FogD. Soap lather |
| Answer» Correct Answer - B::C | |
| 1188. |
A solution sontaining 10.2 g of elycring pe litre is found to be isotonic with 2% solution of glucose `("molar mass"=180 g mol^(-1))`. Calculate the molar mass of glycrine. |
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Answer» Since solution are isotonic ,`pi_("glycerine")=pi_("glucose")` `pi_("glycerine")=(W_(B)xxRxxT)/(M_(B)xxV)` `W_(B)=10.2g, V=1 L` `pi_("glycerine")=((10.2g)xxRxxT)/((M_(B))xx(1L))` `pi_("glucose")=(W_(B)xxRxxT)/(M_(B)xxV)` ` W_(B)=2g, M_(B)=(180 gmol^(-1),V=100 mL=0.1L` `pi("glucose")=((2g)xxRxxT)/((180g mol^(-1))xx(0.1L))` Equating eqns, (I) and (II): ` ((10.2g)xxRxxT)/((M_(B))xx(1L))=((2g)xxRxxT)/((180g mol^(-1))xx(0.1L))` `M_(B)=((10.2g)xx(180g mol^(-1))xx(0.1L))/((1L)xx(2g))=91.8 g mol^(-1)` Molar mass of glycerisng= 91.8 g `mol^(-1)` |
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| 1189. |
102% solution of glycerine and 2% solution of glucose are isotonic. Molecular mass of glucose is 180 then find out the molecular mass of glycerine. |
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Answer» `pi_("glycerine")=pi_("glucose")` `:. C_(("glycerine"))=C_(("glucose"))` `=(1.02xx1000)/(m_("glycerine")xx100)=(2xx1000)/(180xx100)` `m_("glycerine")=91.8`. |
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| 1190. |
The solution which boil at constant temperature like a pure liquid and possess same composition in liquid as well as vapour state are called azeotropes. The components of azetropes cannot be separated by fractional distillation. Only non-ideal solutions form azeotropes. Solutions with negative deviation form maximum boiling azeotrope and the solutions with positive deviation form minimum boiling azeotrope. The boiling point of an azeotrope is never equal to the boiling points of any of the components of the azeotrope. Answer the following question: The azeotropic solutions of two miscible liquids Solutions which distill without any change in composition or temperature are calledA. SaturatedB. SupersaturatedC. IdealD. Azeotrope |
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Answer» Correct Answer - D Anetropes behave as pure solution, therefore they boil at a fixed temperature and hence they cannot be distilled off. |
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| 1191. |
The solution which boil at constant temperature like a pure liquid and possess same composition in liquid as well as vapour state are called azeotropes. The components of azetropes cannot be separated by fractional distillation. Only non-ideal solutions form azeotropes. Solutions with negative deviation form maximum boiling azeotrope and the solutions with positive deviation form minimum boiling azeotrope. The boiling point of an azeotrope is never equal to the boiling points of any of the components of the azeotrope. Answer the following question: The azeotropic solutions of two miscible liquids The azeotropic mixture of water and `HCl` boils at `108.5^(@)C`. This solution isA. IdealB. Non-ideal with positive deviationC. Non-ideal with negative deviationD. None |
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Answer» Correct Answer - C Azetropes are formed only by non-ideal solutions. |
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| 1192. |
An aqueous solution containing 288gm of a non-volatile compound having the stochiometric composition `"C"_("X")"H"_("2X")"O"_("X")` in 90gm water boils at `101.24^(@)"C"` at 1.00 atmospheric pressure. What is the molecular formula? `"K"_("b")("H"_(2)"O")=0.512" K mol"^(-1) "kg"" T"_("b")("H"_(2)"O")=100^(@)"C"` |
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Answer» Elevation in B.P `=101.24-100=1.24^(@)"C"` `Delta "T"_("b")="K"_("b")xx"i"xx"molality" implies1.24=0.512xx288/"m"xx1000/90 (therefore"i"=1)` `therefore "m"=1321.2"gm mol"^(-1)` molar mass of `"C"_("X")"H"_("2X")"O"_("X")=12"x"+1xx2"x"+16"x"=30"x"` `therefore "x"=44` Hence the molecular formula is `="C"_("44")"H"_("88")"O"_("44")` |
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| 1193. |
Two liquids say X and Y boil at 380 K and 400 K respectively. Which of them is more volatile? Why? |
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Answer» Liquid – X because lower the boiling point more will be volatile (evaporation)6. |
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| 1194. |
Which is the correct statement ?A. Minimum boiling azeotropic mixture boils at a temperature higher than either of the two componentsB. Maximum boiling azeotropic mixture boils at temperature lower than either of the two componentsC. Minimum boiling azeotropic mixture shows positive (+ve) deviationD. Maximum boiling azeotropic mixture shows nagative (-ve) deviation. |
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Answer» Correct Answer - (c,d) are both correct op[tions. |
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| 1195. |
What is the effect of rise in temperature on solubility of gases? |
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Answer» Dissolution of gas is exothermic process. Hence according to Le- Chatelier‘s principle, the solubility of gas should decrease with rise in temperature. |
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| 1196. |
Assertion : Molarity of a solution in liquid state changes with temperatuer. Reson : The volume of a solution changes with change in temperature.A. Assertion and reson both are correct statements and reson is sorrect explanation for assertion.B. Assertion and reson both are correct statements but resons is not correct explanation for assertion.C. Assertion and reason both are incorrect statements.D. Assertion is wrong statement but reason is correct statement. |
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Answer» Correct Answer - a Reason is the correct explanation for assertion. |
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| 1197. |
What happens to the solubility of a gas in a solution with rise in temperature ? |
| Answer» Correct Answer - It decreases. | |
| 1198. |
Which of the following binary mixture will have same composition in liquid and vapour phase?A. Benzene-tolueneB. Water-nitric acidC. Water-ethanolD. n-hexane-n-heptane |
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Answer» Correct Answer - B::C Mixture having same composition in liquid and vapour phase are known as azeotropes. Azeotropes boils at same temperature. Here, water-nitric acid water-entanol mixtures are non-ideal solution. Hence, water-nitric acid and water-ethanol ae examples of azeotropes. While benzene-toluene and n-hexane -n heptanc are examples of ideal solution |
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| 1199. |
To `250 mL` of water, `x g` of acetic acid is added. If `11.5%` of acetic acid is dissociated, the depressin in freezing point comes out `0.416`. What will be the value of `x` if `K_(f) ("water")=1.86 K kg^(-1)` and density of water is `0.997 g mL`. |
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Answer» Given that: `W_(2)=x g` ,`DeltaT_(f)=0.416`,`alpha=0.115` `alpha=(i-1)/(n-1)` So, `i=1.115` `DeltaT_(f)=ixxmmxxK_(f)` `DeltaT_(f)=ixx(W_(2)xx1000xxK_(f))/(Mw_(2)xxW_(1))` `0.416=1.115xx(x xx 1000 xx1.86)/(60xx249.25)` `[W_(2)= x g]` x=`(0.416xx60xx249.25)/(1.115xx1000xx1.86)=3 g` |
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| 1200. |
Normal boiling point `(T_(N))` is defined as the temperature when V.P. of liquid become equal to `1`atm and standard boiling point `(T_(S))` is defind becomes equal to `1`bar . Which one is not correct .if water is considered?A. `T_(N)=100^(@)C`B. `T_(S) gt 100^(@)C`C. `T_(S) lt 100^(@)C`D. `T_(S) lt T_(N)` |
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Answer» Correct Answer - B `T_(S)=372.6K` for water since `1` bar `lt 1` `(1 "bar" = 0.998 "atm")`. |
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