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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.
| 151. |
Solubility of deliquescent substances in water is generally:A. HighB. LowC. ModerateD. Cannot be said |
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Answer» Correct Answer - A A deliquescent solid is one which absorbs so much amount of water that if forms a saturated solution of it. |
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| 152. |
A substance will be deliquescent if its vapour pressure is:A. (a) Equal to the atmospheric pressureB. (b) Equal to that of water vapour in the airC. (c ) Greater than that of water vapour in the airD. (d) Lesser than that of water vapour in the air |
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Answer» Correct Answer - D Higher vapour pressure of `H_(2)O` in atmosphere will derive `H_(2)O` vapours to solute particles. |
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| 153. |
A supersaturates solution is a metastable state of solution in which solute concentration:A. (a) Is equal to the solubility of that substance in waterB. (b) Exceeds than its solubilityC. (c ) Less than its solubilityD. (d) Continuously change |
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Answer» Correct Answer - B It is the characteristic of supersaturated solution the meta stable state leading to saturated solution after few time. |
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| 154. |
Two beaker A and B present in a closed vessel. Beaker A contains 152.4 g aqueous soulution of urea, containing 12 g of urea. Beaker B contain 196.2 g glucose solution, containing 18 g of glocose. Both solution allowed to attain the equilibrium. Determine mass % of glocose in its solution at equilibrium allowed to attain the equilibrium :A. 6.71B. 14.49C. 16.94D. 20 |
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Answer» Correct Answer - b |
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| 155. |
The freezing point of a `0.05` molal solution of a non-electrolyte in water is: (`K_(f) = 1.86 "molality"^(-1)`)A. (a) `-1.86^(@)C`B. (b) `-0.93^(@)C`C. (c ) `-0.093^(@)C`D. (d) `0.093^(@)C` |
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Answer» Correct Answer - C `Delta T_(f) = 1.86 xx 0.05 = 0.093`, `:. T_(f) = 0-0.093 = -0.093^(@)C` |
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| 156. |
A maximum or minima obtained in the temperature, composition curve of a mixture of two liquids indicates:A. (a) An azeotropic mixtureB. (b) An eutectic formationC. ( c) That the liquids are immiscible with one anotherD. (d) That the liquids are partially miscible at the maximum or minimum |
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Answer» Correct Answer - A A characteristic of azeotropes. |
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| 157. |
A mixture of two immiscible liquids nitrobenzene and water boiling at `99^(circ)C` shows a partial pressure of the closest value of their weight ratio. (Water: nitrobenzene) |
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Answer» Correct Answer - 4 `27 = (760 xx (w_(2))/(123))/((w_(1))/(18) + (w_(2))/(123))` `733 = (760 xx (w_(1))/(18))/((w_(1))/(18)+(w_(2))/(123))` or `(w_(1))/(w_(2)) = 4` |
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| 158. |
A compound has the empirical formula `C_(10)H_(8)Fe`. A solution of 0.26 g of the compound in 11.2 g of benzene (`C_(6)H_(6)`)boils at `80.26^(@)C`. The boiling point of benzene is `80.10^(@)C`, the` K_(b)` is `2.53^(@)C`/molal. What is the molecules formula of the compound?A. `C_(30)H_(24)Fe_(3)`B. `C_(10)H_(8)Fe`C. `C_(5)H_(4)Fe`D. `C_(20)H_(16)Fe_(2)` |
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Answer» Correct Answer - d |
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| 159. |
Water and chorobenzene are immiscible liquids. Their mixture boils at `89^(@)C` is 7 x`10^(4)`pa. Mass per cent of chorobenzene in the distilate is :A. 50B. 60C. 78.3D. 38.46 |
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Answer» Correct Answer - d |
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| 160. |
Azeotropic mixture of water and `C_(2)H_(5)OH` boils at 351 K. By distilling the mixture it is possible to obtainA. pure `C_(2)H_(5)OH ` onlyB. Pure water onlyC. Neither`C_(2)H_(5)OH` nor waterD. Both water and `C_(2)H_(5)OH` in pure state |
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Answer» Correct Answer - c |
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| 161. |
The azeotropic mixture of water (`B.P. = 100^(@)C`) and HCl(`B.P. =86^(@)C`)boils at about `120^(@)C`. During fractional distillation of this mixture it is possible to obtain :A. pure HClB. pure`H_(2)O`C. pure `H_(2)O` as well as pure HClD. Neither` C_(2)H_(5)OH` nor HCl |
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Answer» Correct Answer - d |
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| 162. |
At `48^(@)C`, the vapour pressure of pure `CS_(2)` is 850torr . A solution of 2.0 g of sulphur in 100g of `CS_(2)` has a vapour pressure 844.9 torr. Determine the atomicity of sulphur molecule :A. 1B. 2C. 4D. 8 |
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Answer» Correct Answer - d |
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| 163. |
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` ?A. 0.1 gB. 1.24 gC. 0.48 gD. 4.8 g |
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Answer» Correct Answer - b |
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| 164. |
When non-volatile solute is assed to a pure solvent, the:A. vapour pressure of the solution becomes lower than the vapour pressure of the pure solventB. rate of evaporation of solvent is reducedC. solute does not affect the rate of condensationD. none of these |
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Answer» Correct Answer - a,b,c ANSWER -: A,B,C) AS ON DISSOLVING NON VOLATILE SOLUTE IT LOWERS DOWN THE VAPOUR PRESSURE OF SOLUTION Ps = P•A XA WHERE A STANDS FOR SOLVENT. |
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| 165. |
In which of the following molecular weight determination methods, sensitivity of the measurements decreases as the molecular weight of the solute increases?A. (a) Elevation of boiling point`//`depression in f.pt.B. (b) ViscosityC. (c ) Osmotic pressureD. (d) None of these |
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Answer» Correct Answer - A `Delta T = (k_(b) xx 1000 xx w)/(m xx W)`, if `m` is more, `Delta T` will be small and thus sensitivity to read `Delta T` accurately decreases. |
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| 166. |
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. (a) `31.25`B. (b) `3.125`C. (c ) `312.5`D. (d) None of these |
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Answer» Correct Answer - A `P^(@)-P_(S) prop` loss in weight of water chamber and `P_(S) prop` loss in `wt`. of solution chamber. `(P^(@)-P_(S))/(P^(@)) = (n)/(N) = (w xx M)/(m xx W)` or `(0.04)/(2.50) = (5 xx 18)/(m xx 180)` `:. m = 31.25` |
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| 167. |
if the elevation in boiling point of a solution of non-volatile, non-electrolytic and non-associanting solute in solvent (`K_(b) =xK.kgt."mol"^(-1)`)is yK,then the depression in freezing point of solution of same concentration would be (`K_(f)` )of the sovent = `zk.kg"mol"^(-1)`)A. `2xy/y`B. `yz/x`C. `xz/y`D. `yz/(2x)` |
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Answer» Correct Answer - b |
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| 168. |
A solution contaning 30 g of a non-volatile solute exactly in 90 g of water has a water has a vapour pressure of 2.8 K Pa at 298 K. Further 18 g of water is added to the solution and the new vapour pressure become 2.9 Pa at 298 K. Calculate. Molecular mass of the solute. Vapour presure of water of water at 2198 K. |
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Answer» `(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W)` For I case: `(P^(@)-2.8)/(2.8) = (30 xx 18)/(m xx 90) = (6)/(m)` …(1) For II case: `(P^(@)-2.9)/(2.9) = (30 xx 18)/(m xx 108) = (5)/(m)` …(2) By eqs. (1) and (2), `P^(@) = 3.53 kPa` `m = 23 g mol^(-1)` Note: Answers will be `3.4 kPa` and `34 g mol^(-1)` if `(P^(@)-P_(S))/(P^(@)) = (n)/(N)` is used which is only valid for dilute solutions. |
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| 169. |
Dry air was suvvessively passed through a solution of `5 g` solute in `80 g` water and then through pure water. The loss in weight of solution was `2.5 g` and that of pure water was `0.04 g`. What is mol.wt. of solute? |
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Answer» Loss in weight of solution `prop P_(S)` Loss in weight of solvent `prop P^(@)-P_(S)` `:. (P^(@)-P_(S))/(P_(S))=("Loss in wt. of solvent")/("Loss in wt. of solution") = (0.04)/(2.5)` …(1) Also, `(P^(@)-(P_(S)))/(P_(S)) = (w xx M)/(m xx W)` …(2) `:. (0.04)/(2.5) = (5 xx 18)/(80 xx m)` or `m = 70.31` |
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| 170. |
(a) `10 g` of a certain non-volatile solute were dissolved in `100 g` water at `20^(@) C`. The vapour pressure was lowered from `17.3555 mm` to `17.2350 mm`, calculate `m`. wt. of solute. (b) The vapour pressure of pure water at `25^(@)C` is `23.62 mm`. What will be the vapour pressure of a solution of `1.5 g` of urea in `50 g` of water? |
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Answer» (a) `(P^(@)-P_(S))/(P_(S)) = (n)/(N) = (w)/(m) xx (M)/(W)` Given that, `P^(@) = 17.3555 mm, P_(S) = 17.2350 mm`, `w = 10 g, W = 100g, M = 18` `(17.3555-17.2350)/(17.2350) = (10 xx 18)/(m xx 100)` `:. m = 257.45` (b) `(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W)` `(23.62-P_(S))/(P_(S)) = (1.5 xx 18)/(60 xx 50)` `:. P_(S) = 23.41 mm` |
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| 171. |
If the boiling point of an aqueous solution containing a non-volatile solute is `100.15^(@)C`. What is its freezing point? Given latent heat of fusion and vapourization of water `80 cal g^(-1)` and `540 cal g^(-1)`, respectively.A. `0.361^(@)C`B. `-0.361^(@)C`C. `-3.61^(@)C`D. None of these |
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Answer» Correct Answer - b |
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| 172. |
Find the molality of a solution containing a non-volatile solute if the vapour pressure is `2%` below the vapour pressure or pure water. |
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Answer» Given, `P_(S) = (98)/(100)P^(@)` Now `(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W)` `= (w)/(m xx W) xx 1000 xx (M)/(1000)` `{(P^(@) - (98)/(100).P^(@))/((98)/(100)P^(@))} = "Molality xx (18)/(1000)` `:. Molatity={(2P^(@))/(100xx98/100P^(@))}xx1000/18=1.1334` |
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| 173. |
A certain non-volatile electrolyte contain 40% carbon, 6.7% hydrogen and 53.3% oxygen.An aqueous solution containing 5% by mass of the solute boils at `100.15^(@)`C. Determine molecular formula of the compound(`K_(b) =0.51^(@)C//m`):A. HCHOB. `CH_(3)OH`C. `C_(2)H_(5)OH`D. `C_(6)H_(12)O_(6)` |
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Answer» Correct Answer - d |
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| 174. |
A 5% (w/V ) solution of cane sugar (molecular mass = 342) is isotonic with 1% (w/V) solution of a subtance X. The molecular mass of X is :A. 34.2B. 171.2C. 68.4D. 136.8 |
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Answer» Correct Answer - c |
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| 175. |
A solution containing 4.0 g of PVcC in 2 liter of dioxane (industrial solvent ) was found to have an osmotic pressure 3.0 x `10^(-4)`atm at `27^(@)C`. The molar mass of the polymer (g/mol) will be :A. 1.6 x `10^(4)`B. 1.6 x `10^(5)`C. 1.6 x `10^(3)`D. 1.6 x `10^(2)` |
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Answer» Correct Answer - b |
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| 176. |
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` |
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Answer» Correct Answer - 8 `because pi prop T` `:. (pi_(1))/(T_(1)) = (pi_(2))/(T_(2)) rArr (4)/(273) = (pi_(2))/(546) rArr pi_(2) = 8 atm` |
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| 177. |
At `27^(@)C`, a `5%` solution `("wt."//"vol")` of cane-sugar is isotonic with `8.77 g//litre` of urea solution. Find `m.wt`. of urea, if `m.wt`. of sugar is `342`. Also report the osmotic pressure of solution is `100 mL` each are mixed at `27^(@)C`. |
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Answer» For isotonic solutions, having neither dissociation nor association nature of solutes. `C_(1) = C_(2)` or `(w_(1))/(m_(1)V_(1)) = (w_(2))/(m_(2)V_(2))` For sugar, For urea `(5)/((342 xx 100)/(1000)) = (8.77)/((m xx 1000)/(1000))` `:. m = (8.77 xx 342)/(5 xx 10) = 59.99` On mixing `100 mL` of cane sugar with `100 mL` urea solution, the total volume ow contains `200 mL` in which `5 g` cane sugar and `0.877 g` urea is present. Thus, `pi xx (200)/(1000) = [(5)/(342)+(0.877)/(60)] xx 0.0821 xx 300` `pi = 3.60 atm` |
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| 178. |
At `17^(@)C` the osmotic pressure of an aqueous of sucrose is `1.855 atm` per `150 mL` solution. Calculate the weight of sucrose in solution. |
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Answer» Correct Answer - 4 `pi V = nST = (w)/(m)ST` `pi = 1.855 atm`., `V = (150)/(1000) = 0.15 litre`, `w = ?` `T = 17+273 = 290K, m = 342`, `S = 0.0821 "litre atm" K^(-1) mol^(-1)` `:. W = (pi Vm)/(ST) = (1.855 xx 0.15)/(0.0821 xx 290) ~~ 4 g` |
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| 179. |
A solution of protein (extracted from carbs) was prepared by dissolving `0.75 g` in `125 cm^(3)` of an aqueous solution. At `4^(@)C` and osmotic pressure rise of `2.6 mm` of the solution was observed. Then molecular weight of protein is (assume density of solution is `1.00 g//cm^(3)`):A. (a) `9.4 xx 10^(5)`B. b) `5.4 xx 10^(5)`C. ( c) `5.4 xx 10^(10)`D. (d) `9.4 xx 10^(10)` |
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Answer» Correct Answer - B Given `h = 2.6 mm` `:. pi = hdg = (2.6)/(10) xx 1 xx 980 " dyne " cm^(-2)` Also `pi = (w)/(V.m)ST` (in `C.G.S`. System) or `(2.6 xx 1 xx 980)/(10) = (0.75 xx 8.314 xx 10^(7) xx 277)/(125 xx m)` `:. m = 5.4 xx 10^(5)` |
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| 180. |
At `20^(@)C`, the osmotic pressure of urea solution is `400 mm`. The solution is diluted and the temperature is raised to `35^(@)C`, when the osmotic pressureis found to be `105.3 mm`. Determine extent of dilution. |
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Answer» For initial solution, `pi = (400)/(760)atm, T = 293 K` `(400)/(760) xx V_(1) = n xx S xx 293` …(i) After dilution, let volume becomes `V_(2)` and temperature is raised to `35^(@)C`, i.e., `308 K`. and `pi = (105.3)/(760)atm` `(105.3)/(760) xx V_(2) = n xx S xx 308` ...(2) By eqs. (1) and (2), we get `(V_(1))/(V_(2)) = (293)/(308) xx (105.3)/(400)` `(V_(1))/(V_(2)) = (1)/(4)` or `V_(2) = 4V_(1)` i.e., Solution was diluted to `4` times. |
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| 181. |
Calculate the osmotic pressure at `17^(@)C` of an aqueous solution containing `1.75 g` of sucrose per `150 mL` solution. |
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Answer» Given that, `w = 1.75 g`, `m = 342` `V = (150)/(1000)litre`, `T = 290 K` `piV = nST = (w)/(m)ST` `pi xx (150)/(1000) = (1.75)/(342) xx 0.0821 xx 290` `pi = 0.812 atm` |
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| 182. |
The osmotic pressure of urea solution is `500 mm` at `10^(@)C`. The solution is diluted and the temperature raised to `25^(@)C`, when the osmotic pressure is found to be `105.3 mm`. Find the extend if dilution. |
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Answer» Correct Answer - 5 `because pi V = nST` Initially `pi_(1) = (500)/(760)atm, T_(1) = 283K` `(500)/(760) xx V_(1) = nS xx 283` …(1) Let on dilution, the volume become `V_(2)` and temperature raise to `25^(@)C` i.e., `298K` `pi_(2) = (105.3)/(760)` Now `(105.3)/(760) xx V_(2) = nS xx 298` ...(2) By eq. (1) and (2) `(V_(1))/(V_(2)) = (283)/(298) xx (105.3)/(500)` `:. V_(1)//V_(2) = 1//5` or `V_(2) = 5V_(1)` `:.` Solution is `5` times diluted. |
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| 183. |
At `25^(@)C`, a solution containing `0.2 g` of polyisobutylene in `100 mL` of benzene developed a rise of `2.4 mm` at osmotic equilibrium. Calculate the molecular weight of polyisobutylene if the density of solution is `0.88 g// mL`. |
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Answer» Height developed `= 2.4 mm` Osmotic pressure `= h.d.g` `= (2.4)/(10) xx 0.88 xx 981` `= 207.187 "dyne" cm^(-2)` Now `piV = nST` `207.187 xx 100 = (0.2)/(m) xx 8.314 xx 10^(7) xx 298` (`R` in erg, `V` in `mL`, using `CGS` system) `m = 2.39 xx 10^(5)` |
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| 184. |
Two aqueous solutions, A and B, are separated by a semi- permeable membrane. The osmotic pressure of solution A immediately begins to decrease. Which of the following statement is ture ?A. The solvent molecular are moving from the solution of higher osmotic pressure to that of lower osmotic pressureB. The initial osmotic pressure of solution B is greater than that of solution A.C. Solvent molecules are moving from solution B into solution A.D. Both (a) and (b) are ture statements. |
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Answer» Correct Answer - c |
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| 185. |
The cryoscopic contant of water is 1.86 K kg `mol^(-1)`. A 0.01 molal acetic acid solution produces a depression of `0.0194^(@)C` in the freezing point. The degree of dissociation of acetic acid is :A. zeroB. 0.043C. 0.43D. 1 |
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Answer» Correct Answer - b |
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| 186. |
In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be (`K_(f)` = 1.86 K kg `mol^(-1)`):A. 274.674 KB. 271.60 KC. 273 KD. none of these |
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Answer» Correct Answer - b |
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| 187. |
Calculate depression of freezing point for 0.56 molal aq. Solution of KCl. (Given : `K_f(H_(2)O) = 1.8 kg mol^(-1)`). |
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Answer» Correct Answer - 2 |
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