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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.
| 101. |
Each question contains STATEMENT-I(Assertion) and STATEMENT-2(Reason).the statement carefully and mark the correct answer accoring to the instrution given below: STATEMENT - 1 : if a liquid solute more volatile than the solvent is added to the solvent, the vapour pressure of the solution is greater than vapour pressure of pure solvent. STATEMENT - 2 : Vapour pressure of solution is eqeal to vapour pressure of sovent.A. If both the statements are TURE and STATEMENT-2 is the correct explanation STATEMENT-1B. If both the statements are TURE but STATEMENT-2 is NOT the correct explanation STATEMENT-1C. If STATEMENT-1 is TURE and STATEMENT-2 is FALSED. If STATEMENT-1 is FA,SE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - C |
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| 102. |
Equal valumes of 0.1 M urea and 0.1 M glucose are mixed. The mixture will have :A. lower osmotic pressureB. same osmotic pressureC. higher osmotic pressureD. none of these |
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Answer» Correct Answer - b |
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| 103. |
If 0.1 M `H_(2)SO_(4)`(aq.) solution shows freezing point `-0.3906^(@)C` then what is the `K_(a2)"for"H_(2)SO_(4) `? (Assume m = M and `K_(f(H_(2)O) = 1.86 K kg mol^(-1)`)A. 0.122B. 0.0122C. 1.11x `10^(-3)`D. None of these |
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Answer» Correct Answer - b |
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| 104. |
Calculate the amount of ice that will separate out on cooling containing `50 g `of ethylene glycol in `200 g` of water to `-9.3^(@)C (K_(f)` for water =`1.86 K mol^(-1) kg`) |
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Answer» Given `DeltaT = 9.3, w = 50 g`, `K_(f)` for `H_(2)O = 1.86 K mol^(-1) kg, m_("glycol") = 62` `because Delta T = (1000 K_(f) xx w)/( m xx W)` `:. 9.3 = (1000 xx 1.86 xx 50)/(62 xx W)` `:. W_("water") = 161. 29 g` Thus, weight of ice seprated `= 200-161.29` `= 38.71 g` |
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| 105. |
A 1.0 g sample of co`(NH_(2)CH_(2)CH_(2)NH_(2))_(3)Cl_(3)` is dissocialved in 25.0 g if water and the freezing point of the solution is `-0.87^(@)C`. How many ions are produced per mole of compound? The `K_(f)` of water is `1.86^(@)C//"molal"`A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - c |
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| 106. |
`C_(6)H_(6)` freezes at `5.5^(@)C`. At what tempreature will a solution of 10.44 g of `C_(4)H_(10)` in 200 g of `C_(6)H_(6) "freeze" K_(f)(C_(6)H_(6))= 5.12^(@)C//m`A. `4.608^(@)C`B. `0.892^(@)C`C. `5.5^(@)C`D. none of these |
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Answer» Correct Answer - b |
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| 107. |
If 1m solution of benzoic acid in benzene has a freezing point depression of `3.84 ^(@)C` . `(K_(f)=5.12^(@)C mol^(-1)`kg) and boiling point elevation of `2.53^(@)C(K_(b)` =`2.53^(@)C "mol"^(-1) kg)`, then select the correct statement/s :lt brgtstatement I : there are dimar formation when under =going freezing Statement II : there are no change when undergoing boiling Statement III : reverse of I and II ltbr. Statement IV : dimer formation in freezing and boiling stateA. I, IIB. II, IIIC. III, ID. only I |
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Answer» Correct Answer - a |
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| 108. |
In the depression of freezing point experiment, it is found that the:A. `V.P.` of the solution is less than that of pure solventB. `V.P.` os the solution is more than that of pure solventC. only solute molecules solidify at the freezing pointD. only solvent molecules solidify at the freezing point |
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Answer» Correct Answer - A::D Both are fact. |
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| 109. |
For different aqueous solutions of `0.1N` urea, `0.1N NaCl, 0.1N Na_(2)SO_(4)` and `0.1N Na_(3)PO_(4)` solution at `27^(@)C`, select the correct statements:A. The order of osmotic pressure is: `NaCl gt Na_(2)SO_(4) gt Na_(3)PO_(4) gt "urea"`B. `pi = (Delta T_(b))/(K_(b)) xx ST` for urea solutionC. Addition of salt on ice increases its melting pointD. Addition of salt on ice brings in melting of ice earlier |
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Answer» Correct Answer - A::B::D Molar concentration are `0.1 M` urea, `0.1 M NaCl`, `(0.1)/(2) M Na_(2)SO_(4)` and `(0.1)/(3)M Na_(3)PO_(4)` `:. pi prop CX (1-alpha +Xalpha +Y alpha) prop C(X+Y)`, if `alpha = 1` `:. pi_("urea") prop 0.1 xx 1, pi_(Na_(3)PO_(4)) prop (0.1)/(3) xx 4` Also `pi = C_(M) xx S xx T` and `Delta T_(b) = "Molality" xx K_(b)` If Molality `=` Molality (for dilute solution) `pi = (Delta T_(b))/(K_(b)) xx S xx T` Also addition of salt on ice lowers its melting point and thus, ice melts earlier. |
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| 110. |
At higher altitude, boiling of water is `95^(@)C`. The amount of `NaCl` added to `1 kg` water `(k_(b)=0.52 K mol^(-1)kg)` in order to raise the b.pt. of solution to `100 ^(@)C` is (assume `90%` ionisation of `NaCl`):A. `296.5 g`B. `281.25 g`C. `270 g`D. `310 g` |
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Answer» Correct Answer - A `Delta T = (1000 xx K_(b) xx w)/(m xx W)(1+alpha)` `5 = (1000 xx 0.52 xx w)/(58.5 xx 1000) xx (1.90)` , `(because alpha = 0.9)` |
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| 111. |
On addition of a volatile liquid `A` to another volatile liquid `B` in any proportiona will always………the vapour pressure of `B`:A. (a) DecreaseB. (b) IncreaseC. (c ) Increase or decreaseD. (d) None of these |
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Answer» Correct Answer - C `V.P.` of mixture lies in between vapour pressure of two liquids. |
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| 112. |
Shrinking of graphs in conc. `NaCl` solution is due to:A. (a) ExosmosisB. (b) EndosmosisC. (c ) Both (a) and (b)D. (d) None of these |
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Answer» Correct Answer - A Solvent particles flow from graphs walls to external solution. |
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| 113. |
The colligative properties are……… proportional to each other:A. (a) InverselyB. (b) DirectlyC. (c ) Both (a) and (b)D. (d) None of these |
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Answer» Correct Answer - B `pi prop` molality, for dilute solutions, `Delta T prop` molality `prop` molarity. Thus `pi prop Delta T`. |
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| 114. |
For exact determination of molecular mass through colligative properties measurement :A. solute must be volatileB. solution must be vary diluteC. solution must be formed by similar nature of subtancesD. solute must not be dissociated |
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Answer» Correct Answer - b,d |
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| 115. |
Statement Osmosis is a bilateral process. Explanation In osmosis net flow from dilute to concentrated solution is noticed.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Explanation is correct reason for statement. |
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| 116. |
Based upon the technique of reverse osmosis the approximate pressure required to desalinate sea water containing 2.5% (mass/volume)`KNO_(3)` at `27^(@)`C will beA. 10.5 atmB. 21 atmC. 12.2 atmD. 6.09 atm |
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Answer» Correct Answer - c |
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| 117. |
Statement- Reverse osmosis is used to purify saline water. Explanation- Solvent molecules pass from concentrate solution to dilute solution through semipermeable membrane if high pressure is applied on solution side.A. If both the statements are TURE and STATEMENT-2 is the correct explanation STATEMENT-1B. If both the statements are TURE but STATEMENT-2 is NOT the correct explanation STATEMENT-1C. If STATEMENT-1 is TURE and STATEMENT-2 is FALSED. If STATEMENT-1 is FA,SE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - B |
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| 118. |
Which of the following is correct for a non-ideal solution of liquids A and B showing negative deviation?A. `/_H_("mix") = - ve`B. `/_V_("mix") = - ve`C. `/_S_("mix") = - ve`D. `/_G_("mix") = - ve` |
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Answer» Correct Answer - a,b,d |
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| 119. |
At `80^@C`, the vapour pressure of pure liquid `A` is `520 mm` Hg and that of pure liquid `B` is `1000 mm Hg`. If a mixture of solution `A` and `B` boils at `[email protected]` and `1 atm` pressure, the amount of `A` in the mixture is `(1 atm =760 mm Hg)` a. `50 mol %` , b.`52 mol %` ,c.`34 mol%` ,d.`48 mol %`A. `52` mol per centB. `34` mol per centC. `48` mol per centD. `50` mol per cent |
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Answer» Correct Answer - D `P_(m) = P_(A)^(@)X_(A) + P_(B)^(@)X_(B)` `760 = 520 X_(A) + P_(B)^(@)(1-X_(A))` `:. X_(A) = 0.5 = 50%` |
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| 120. |
At `40^(@)C`, the vapour pressure in torr of methyl and ethyl alcohol solutions is represented by `P = 119 X_(A)+135`, where `X_(A)` is mole fraction of methyl alcohol. The value of `(P_(B)^(@))/(X_(B))` at lim `X_(A) rarr 0)`, and `(P_(A)^(@))/(X_(A))` at lim `X_(B) rarr 0` are:A. (a) `135, 254`B. (b) `135, 230`C. ( c) `119, 135`D. (d) `140, 135` |
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Answer» Correct Answer - A when `X_(A)rarr0, X_(B)rarr1`, Thus, `(P_(B)^(@))/(X_(B))=135 and (P_(A)^(@))/(X_(A))=254` |
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| 121. |
The vapour pressure of benzene at `90^(@)C` is `1020` torr. A solution of `5 g` of a solute in `58.5 g` benzene has vapour pressure `990` torr. The molecualr weight of the solute is:A. (a) `78.2`B. (b) `178.2`C. ( c) `206.2`D. (d) `220` |
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Answer» Correct Answer - D `(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W)` `(1020 - 990)/(990) = (5 xx 78)/(m xx 58.5)` `:. m = 220` |
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| 122. |
At `310 K`, the vapour pressure of an ideal solution containing `2 "moles"` of `A` and `3"moles"` of `B` is `550 mm` of `Hg`. At the same temperature, if one mole of `B` is added to this solution, the vapour pressure of solution increase by `10 mm of Hg`. Calculate the `V.P` of `A` and `B` in their pure state. |
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Answer» Initially,`P_(M) = P_(A)^(@).X_(A) + P_(B)^(@).X_(B)` `550 = P_(A)^(@).((2)/(2+3))+P_(B)^(@).((3)/(2+3))` or `2P_(A)^(@) + 3P_(B)^(@) = 2750` …(1) When `1` mole of `B` is further added to it, `P_(M) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B)` `560 = P_(A)^(@) ((2)/(2+4)) + P_(B)^(@).((4)/(2+4))` `:. 2P_(A)^(@) + 4P_(B)^(@) = 3360` ...(2) By eqs. (1) and (2), `P_(A)^(@) = 460 mm, P_(B)^(@) = 610 mm` |
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| 123. |
Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to `4` kg for water to prevent it from freezing at `-6^(@)C` will be (`K_(f)` for water =`1.86 K kg mol^(-1)` and molar mass of ethylene glycol =`62 g mol^(-1)`)A. `800.00 g`B. `204.30 g`C. `400.00 g`D. `304.60g` |
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Answer» Correct Answer - A `Delta T_(f) = K_(f) xx "molality" = K_(f) xx (w xx 1000)/(m xx W)` `w = ?, W = 4 xx 10^(3)g, m = 62` `Delta T_(f) = 0-(-6) = 6` `w = (6 xx 62 xx 4 xx 10^(3))/(1000 xx 1.86) = 800 g` |
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| 124. |
The vapour pressure of two liquid P and Q are 80 torr and 60 torr respectively. The total vapour pressure obtained by mixing `3` moles of P and 2 mole of Q would beA. `69=8 "torr"`B. `20 "torr"`C. `140 "torr"`D. `72 "torr"` |
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Answer» Correct Answer - D `P_(M) = P_(1)^(@).X_(1)+P_(2)^(@).X_(2)` `= 80 xx (3)/(3+2)+60 xx (2)/(3+2) = 72 "torr"` |
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| 125. |
Which one of the following pairs of solution can we expect to be isotonic at the same temperatureA. 0.1 M urea and 0.1 M NaClB. 0.1 M urea and 0.2 M`MgCl_(2)`C. 0.1 M NaCl and 0.1 M `Na_(2)SO_(4)`D. 0.1 M `C(NO_(3))_(2) "and" 0.1 m Na_(2)SO_(4)` |
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Answer» Correct Answer - d |
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| 126. |
One molal solution of a carboxylic acid in benzene shows the elevation of boiling point of 1.518 K. The degree of association for simerization of the acid in benzene is (`K_(b)` for beznene = `2.53 K kg mol^(-1)` ):A. 0.6B. 0.7C. 0.75D. 0.8 |
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Answer» Correct Answer - d |
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| 127. |
Calculate elevation in boiling point for 2 molal aqueous solution of glucose. (Given `K_b(H_(2)O) = 0.5 kg mol^(-1)`) |
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Answer» Correct Answer - 1 |
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| 128. |
Calculate the boiling point of a solution containing `0.61 g` of benzoic acid in `50 g` of carbon disulphide assuming `84%` dimerization of the acid. The boiling point and `K_(b)` of `CS_(2)` are `46.2^(@)C` and `2.3 kg mol^(-1)`. |
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Answer» `Delta T = (1000 xx K_(b) xx w)/( m xx W)` `Delta T_(N) = (1000 xx 2.3 xx 0.61)/(122 xx 50)` `(m`.wt. of `C_(6)H_(5)COOH = 122`) `Delta T_(N) = 0.23` `Delta T_(exp.) = Delta T_(N) (1-alpha+(alpha)/(n)) = 0.23 xx (1-0.84+(0.84)/(2))` `= 0.23 xx (1-(0.84)/(2)) = 0.1334` B. pt. `= 46.2 + 0.1334 = 46.3334^(@)C` |
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| 129. |
Calculate the osmotic pressure of `20%` (wt.`//` vol.) anhydrous `Ca(NO_(3))_(2)` solution at `0^(@)C` assuming `100%` ionisation. |
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Answer» `{:(,Ca(NO_(3))_(2)rarr,Ca^(2+)+,2NO_(3)^(-),),("Before dissociation",1,0,0,),("After dissociation",1-alpha,alpha,2alpha,):}` Given, `w = 20g, V = 100mL, T = 273 K`, Mol, wt. of `Ca(NO_(3))_(2) = 164` `pi_(N) = (w)/(m.V) xx S xx T` `= (20 xx 1000 xx 0.0821 xx 273)/(164 xx 100) = 27.33 atm` Now, `(pi_(exp.))/(pi_(N)) = 1 + 2a = 1+2 = 3` `because alpha = 1` `:. pi_(exp.) =27.33 xx 3 = 82.0 atm` |
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| 130. |
For `[CrCl_(3).xNH_(3)]`, elevation in `b.pt` of one molal solution is triple of one molal aqueous solution of urea. Assuming `100%` ionisation of complex molecule, calculated the value of `x`. |
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Answer» `[CrCl_(3). xNH_(3)]` is complex which on `100%` ionisation gives a complex ion and ionised `Cl^(-)` ions. Since `DeltaT_(b)` complex `=3xxDeltaT_(b)` urea Thus complex molecule should furnish three ions. Therefore complex is `[CrCl.xNH_(3)]Cl_(2) rarr [CrCl.xNH_(3)]+2Cl^(-)` Also co-ordination number of `Cr` is six. thus `x=5` |
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| 131. |
Pure water boils at `99.725^(@)C` at Shimla. If `K_(b)` for water is `0.51 K mol^(-1) kg` the boiling point of `0.69` molal urea solution will be:A. (a) `100.35`B. (b) `100.08`C. (c ) `99.37`D. (d) None of these |
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Answer» Correct Answer - B `DeltaT_(b)=Kxx"molality"=0.51xx0.69=0.352` `:. b.pt.=99.725+0.352-100.077^(@)C` |
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| 132. |
Statement Boiling point of water is `100^(@)C` although water boils below `100^(@)C` on mountains. Explanation Boiling point of a liquid is the temperature at which `V.P.` of liquid becomes equal to `1 atm`.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Water boils at low temperature at mountains where atmoshpheric pressure is low. i.e., when `P^(circ) =` atmoshperic pressure |
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| 133. |
Statement Super heating means to heat a liquid just above its boiling point. Explanation On direct heating, the layer in contact with frame has relatively higher temperature than the other layers of liquid.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - D These are facts about super heating. |
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| 134. |
Statement At equilibrium of Liquid `hArr` vapour, kinetic energy liquid phase and vapour phase are same. Explanation Kinetic energy of liquid or vapour is given by `3//3 RT` for one mole.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Explanation is correct reason for statement. |
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| 135. |
Statement The vapour pressure of `0.1 M` sugar solution is more than that of `0.1 M` potassium chloride solution. Explanation Lowering of vapour pressure is directly proportional to the number of species present in the solution.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - B Statement is wrong. Vapour pressure of `0.1M` sugar is less than `0.1MKCl` solution and explanation is correct. |
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| 136. |
Which represents correct difference when non-volatile solute is present in an ideal solution? A. I, II, IIIB. I, IIIC. II, IIID. I, II |
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Answer» Correct Answer - a |
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| 137. |
Select correct statement :A. Solution has more molecules randomness than a pure solvent. The entropy change between solution and solid is lager than the entropy change between pure solvent and solidB. Heat of fusion of solution are positiveC. Solution containing sugar freezes at a lower tempreature than pure waterD. All are correct statements |
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Answer» Correct Answer - d |
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| 138. |
The vapour pressure of an aqueous solution of glucose is 750 mm of mercury at `100^(@)C`. Calculate the molatlity and mole fraction of solute. |
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Answer» Given that temperature is `372 K` and `b`. Pt. of `H_(2)O = 373 K` `:.` Vapour pressure of `H_(2)O = 76 cm` We have, `(P^(@)-P_(S))/(P_(S)) = ( w xx M)/(mxxW)` `:.` Molality `= (w)/(mxx W) xx 1000 = (P^(@)-P_(S))/(P_(S)) xx (1)/(M) xx 1000` `= (760-750)/(750) xx (1)/(18) xx 1000` `= 0.741 mol//kg` of solvent Also we have, `(P^(@)-P_(S))/(P^(@)) = (n)/(n+N)` `:.` Mole fraction `= (P^(@)-P_(S))/(P^(@)) = (760-750)/(760) = (10)/(760) = 0.013` |
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| 139. |
What weight of glucose (mol.wt. `= 180`) would have to be added to `1700 g` of water at `20^(@)C` to lower its vapour pressure `0.001 mm`? The vapour pressure of pure water is `17 mm Hg` at `20^(@)C`. |
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Answer» Correct Answer - 1 Given, `P^(@) - P_(S) = 0.001 mm`, `P^(@) = 17 mm, P_(S) = 16.999, m = 180`, `M = 18, W = 1700, w = ?` Now `(P^(@)-P_(S))/(P_(S)) = (w//m)/(W//M)` `rArr (0.001)/(16.999) = (w xx 18)/(180 xx 1700) rArr w = 1 g` |
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| 140. |
Glucose is added to `1` litre water to such an extent that `(Delta T_(f))/(K_(f))` becomes equal to `(1)/(1000)`, the weight of glucose added is:A. (a) `0.32 g`B. (b) `0.42 g`C. ( c) `0.22 g`D. (d) `0.18 g` |
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Answer» Correct Answer - D `Delta T_(f) = K_(f) xx "molality" = K_(f) xx (w xx 1000)/(m xx W)` `:. (Delta T_(f))/(K_(f)) = (w xx 1000)/(m xx W)` or `(1)/(1000) = (w xx 1000)/(180 xx 1000)` `:. w = 0.18 g` |
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| 141. |
phenol associates in benzene to a certain extent in dimerisation reaction. A solution containing 0.02 kg of phenol in 1.0 kg of benzene has its freezing point depressed 0.69 k. [`K_(f)(_(6)H_(6)) =5.12 k "mol"^(-1)`]A. 0.63B. 0.73C. 0.83D. 0.93 |
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Answer» Correct Answer - b |
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| 142. |
Which of the following statement(s) is/are correct, if intermolecular forces in liquids A, B and C are in the order of A lt B lt C ?A. B evaporates more readily than AB. B evaporates more readily than CC. A evaporate more readily than CD. all evaporates at same rate at same temperature. |
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Answer» Correct Answer - b,c |
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| 143. |
Lowering in vapour pressure is determined by Ostwald and Walker dynamic methed. It is based on the prinicipal , that when air is allowed to pass through a solvent or solution, it takes up solventvapour with it to get itself saturated at that temperature I and II are weighted separately before and after passing dry air. Loss in mass of each set, gives the lowing of vapour pressure. The temperature of air, the solution and the solvent is kept constant. Loss in masss of solvent (`w_(II)`)will be proportional to :A. `P^(@)-P`B. `P - P^(@)`C. `(P)/(P^(@))`D. `P xx P^(@)` |
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Answer» Correct Answer - a |
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| 144. |
Lowering in vapour pressure is determined by Ostwald and Walker dynamic methed. It is based on the prinicipal , that when air is allowed to pass through a solvent or solution, it takes up solventvapour with it to get itself saturated at that temperature I and II are weighted separately before and after passing dry air. Loss in mass of each set, gives the lowing of vapour pressure. The temperature of air, the solution and the solvent is kept constant. Dry air was passed thorough 9.24 g of solute in 108 g of water and then through pure water. The loss in mass of solution was 3.2 g and that of pure water 0.08 g . The molecular mass (g/mol) of solute is nearly :A. 50B. 62C. 70D. 80 |
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Answer» Correct Answer - b |
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| 145. |
Calcuate the percentage degree of dissociation of an electrolyte `XY_(2)` (Normal molar mass = 164) in water if the water if the observed molar mass by measuring elevation in boiling point is 65.6A. 0.75B. 0.25C. 0.65D. None of these |
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Answer» Correct Answer - a |
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| 146. |
Estimste the lowering of vapour pressure due to the solute (glucose) in a 1.0 M aqueous solution at `100^(@)C`:A. 10 torrB. 18 torrC. 13.45 torrD. 24 torr |
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Answer» Correct Answer - c |
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| 147. |
If the vapor pressure of a dilute aqueous solution of glucose is `750 mm` of Hg at `373 K`, then molality of solute isA. 0.26B. 0.73C. 0.74D. 0.039 |
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Answer» Correct Answer - c |
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| 148. |
The degree of dissociation of `Ca(NO_(3))_(2)` in a dilute aqueous solution containing `7 g` of salt per `100 g` of water at `100^(@)C` is `70%`. Calculate the vapour pressure of solution. |
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Answer» `{:(,Ca(NO_(3))_(2)hArr,Ca^(2+),2NO_(3)^(-),),("Before dissociation",1,0,0,),("After dissociation",1-alpha,alpha,2alpha,):}` `:.` Total mole at equilibrium `=(1+2alpha) ( :. alpha= 0.7)` `=(1+2xx0.7)=2.4` For `Ca(NO_(3))_(2)`: `(m_(N))/(m_(exp))=1+2alpha` `:. m_(exp)=(m_(N))/((1+2xx0.7))=(164)/(2.4)=68.33` Also at `100^(@)C, P_(H_(2)o)^(@)=760mm, w= 7 g, W= 100g` Now `(P^(@)-P_(S))/(P_(S))=(7xx18)/(68.33xx100)=0.0184` or `(P^(@))/(P_(S))-1= 0.0184` `:. P_(S)=(760)/(1.0184)= 746.27 mm` |
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| 149. |
The boiling point of `C_(6)H_(6), CH_(3)OH, C_(6)H_(5)NH_(2)` and `C_(6)H_(5)NO_(2)` are `80^(@)C, 65^@ C ,184^@ C` and `212^(@)C` respectively. Which will show highest vapour pressure at room temerature:A. `C_(6)H_(6)`B. `CH_(3)OH`C. `C_(6)H_(5)NH_(2)`D. `C_(6)H_(5)NO_(2)` |
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Answer» Correct Answer - B Lower is the b.pt. of solvent more is its vapour pressure. |
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| 150. |
The solubility of a salt in water is `40 g` at `30^(@)C`. The amount of water required to dissolve `120 g` at the same temperature is:A. (a) `400 g`B. (b) `4 litre`C. (c ) `300 g`D. (d) `500 g` |
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Answer» Correct Answer - C Solubility is defined as the maximum amount of solute dissolved in `100 g` solvent. |
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