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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. |
25mL of `3M HCI` were added to 75 mL of `0.05M HCI`. The molarity of HCI in the resulting solution is approximatelyA. `0.55M`B. `0.35M`C. `0.787M`D. `3.05M` |
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Answer» Correct Answer - C Molarity of resulting solution `M_("Total") = (M_(A)V_(1))/(V_("Total")) +(M_(2)V_(2))/(V_("Total"))` `= (25 xx 3)/(100) +(75 xx 0.05)/(100)` `= 0.75 +0.0375 = 0.7875M` |
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| 152. |
you are supplied with 500 mL each so 2N HCI and 5N HCI. What is the maximym volume of 3M HCI that you can can prpare using only these two solutions ?A. 250 mLB. 5090 mLC. 750mLD. 1000 mL |
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Answer» Correct Answer - c For solution, consult Example 2,14 (Text part). |
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| 153. |
The vapour pressure to two miscible liquids (A) and (B) are 300 and 500 mm of Hg respectively. In a flask 10 mole of (A) is mixed with 12 mole of (B). However, as soon as (B) is added, (A) starts polymerising into a completely insoluble solid. The polymerisation follows first-order kinetics. After 100 minutes, `0.525` mole of a solute is dissolved which arrests the polymerisation completely. The final vapour pressure of the solution is 400 mm of Hg. Estimate the rate constant of the polymerisation reaction. Assume negligible volume change on mixing and polymerisation and ideal behaviour for the final solution. |
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Answer» Correct Answer - `(1.0 xx 10^(-4))` `P_(T) = P_(A)^(@) x_(A) +P_(B)^(@) x_(B)` `400 = 300 xx ((10-x))/((22.525-x)) +500 xx (12)/((22.525-x))` `9010 - 400 x = 900 - 300 x` `x = 0.1` `K = (1)/(100) In (10)/(10-0.1) = (1)/(100) In (10)/(9.9)` `= 1xx 10^(4) min^(-1)` |
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| 154. |
`2N HCI` solution will have the same molar concentration as aA. `4.0 N H_(2)SO_(4)`B. `0.5 NH_(2)SO_(4)`C. `1N H_(2)SO_(4)`D. `2N H_(2)SO_(4)` |
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Answer» Correct Answer - A `2N HCl=2g eq L^(-1)=2 mol L^(-1)` [Eqn. mass = Mol mass] `4N H_(2)SO_(4)=4g eq L^(-1) = 4xx49 g L^(-1)` `=4xx(49)/(98)=2 mol L^(-1)` (Eq. mass of `H_(2)SO_(4)=49` , Mol mass = 98) |
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| 155. |
The vapour pressure of two miscible liquids `(A)` and `(B)` are `300` and `500 mm` of `Hg` respectively. In a flask `10` mole of `(A)` is mixed with `12` mole of (B). However, as soon as `(B)` is added, `(A)` starts polymerising into a completely insoluble solid. The polymerisation follows first-order kinetics. After `100` minute, `0.525` mole of a solute is dissolved whivh arrests the polymerisation completely. The final vapour pressure of the solution is `400 mm` of `Hg`. Estimate the rate constant of the polymerisation reaction. Assume negligible volume change on mixing and polymerisation and ideal behaviour for the final solution. |
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Answer» Correct Answer - `1.0 xx 10^(-4)` `A+B rarr A_(n)+B` `P_(M)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)` Let a mole of A are left due to polymerization after 100 min. `P_(M) = 300 ((a)/(12+a))+500((12)/(12+a))` ….(i) `k =(2.303)/(100)"log"(10)/(a)` ……(ii) after 100 minute solute is added & final vapour pressure is 400 mm Hg i.e. `P_(s) = 400` `(P_(M)-400)/(400) = (0.525)/((a+12))` ....(iii) from equation (i) and (iii) `a = 9.9` putting this in eq. (ii) k `= (2.303)/(100)"log"(10)/(9.9) = 1.0xx10^(-4)` |
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| 156. |
When the mixture of two immicible liquids (water and nitrobenzene) boils at 372K and the vapour pressure at this temperature are `97.7"k Pa"("H"_(2)"O") "and 3.6 k Pa"("C"_(6)"H"_(5)"NO"_(2)).` Calculate the weight % of nitrobenzene in the vapour. |
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Answer» Correct Answer - 0.2011 Let vapour contains `"W"_(1)" gm of" " C"_(6)"H"_(5)"NO"_(2) and"W"_(2)" gm of"" H"_(2)O"` `"n"_(1)="w"_(1)/123," n"_(1)="w"_(1)/18` for gases Ratio of moles = Ratio of pressure `"n"_(1):"n"_(2)=3.6:97.7` `"w"_(1)/123xx18/"w"_(2)=3.6/97.7` `"w"_(1)/"w"_(2)=(3.6xx123)/(18xx97.7)=0.252` `"w"_(1)="w"("let"),"w"_(1)=0.252"w"` `%"w"_(1)=(0.252"w")/(1.252"w")xx100` `=20.11%` |
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| 157. |
Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300 K when mixed in the molar ratio 1:1. when mixed in the molar ratio of 1:2 at the same temperatre the vapour pressure of the mixture is 350 mm. The vapour pressure of the two pure liquids X and Y respectively areA. `250 mm, 550 mm`B. `350 mm, 450mm`C. `350mm, 700mm`D. `550 mm, 250mm` |
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Answer» In 1st case, `n_(X)=n_(Y)` `therefore x_(X)=0.5` and `x_(Y)=0.5` `therefore p_(s)=p_(X)^(@)x_(X)+p_(Y)^(@)x_(Y)` `400=0.5p^(@)Y=800` ….(i) In 2nd case, `n_(X)//n_(Y)=1//2` `therefore x_(X)=1//3` and `x_(Y)=2//3` `therefore 350=(1//3)p^(@)X +(2//3)p^(@)Y` Or `p^(@)X + 2p^(@)Y = 1050` ....(ii) From (i) and (ii) `p^(@)Y=250` and `p^(@)X=550` |
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| 158. |
Dry air was passed successively through a solution of `5`g of a solute in `180`g water and then through pure water. The loss in weight of solutionwas `250`g and that of pure solvent `0.04`g . The molecular weight of the solute isA. `31.25`B. `3.125`C. `312.5`D. None of these |
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Answer» Correct Answer - A `P_(0)-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)=(wxxM)/(mxxW)` or`(0.04)/(2.50)=(5xx18)/(mxx180)` `:. M=31.25` |
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| 159. |
Two liquids `X` and `Y` are perfectly immisoible. If `X` and `Y` have molecular masses in ratio `1:2`, the total vapour pressure of a mixture of `X` and `Y` prepared in weight ratio `2:3` should be `(P_(X^(0))=400` torr, `P_(Y^(0))=200`torr)A. `600` torrB. `400` torrC. `800` torrD. `1000` torr |
| Answer» Correct Answer - A | |
| 160. |
An ideal solution contains two volatile liquids `A(P^(@)=100` torr) and `B(P^(@)=200` torr). If mixture contain `1` mole of `A` and `4` mole of `B` then total vapour pressure of the distillate is :A. `150`B. `180`C. `188.88`D. `198.88` |
| Answer» Correct Answer - C | |
| 161. |
The vapour pressure of solvent is 20 torr, while that of its dilute solution is 17 torr , the mole-fraction of the solvent isA. `0.6`B. `0.85`C. `0.5`D. `0.7` |
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Answer» Correct Answer - B From question `P^(@)=20` torr `P_(S)=17` torr `:.(P^(@)-P_(s))/(P^(@))`=X solute `(3)/(20)`=X solute or =X solute = `0.15` X solvent =`1-0.15=0.85` |
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| 162. |
For a binary ideal lqiuid solutions,the total pressure of the solution is given asA. `P_("total")=P_(A)^(@)+(P_(A)^(@)-P_(B)^(@))X_(B)`B. `P_("total")=P_(B)^(@)+(P_(A)^(@)-P_(B)^(@))X_(A)`C. `P_("total")=P_(B)^(@)+(P_(B)^(@)-P_(A)^(@))X_(A)`D. `P_("total")=P_(B)^(@)+(P_(B)^(@)-P_(A)^(@))X_(B)` |
| Answer» Correct Answer - B | |
| 163. |
The vapour pressure of solvent is 20 torr, while that of its dilute solution is 17 torr , the mole-fraction of the solvent isA. `0.6`B. `0.4`C. `0.5`D. `0.7` |
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Answer» Correct Answer - B From question `P^(@)=20` torr `P_(S)=17` torr `:. (P^(@)-P_(s))/(P^(@))=X_("solute")` `(3)/(20)=X_(solute")` or `=X_("solute") =0.6` `X_("solvent") =0.4` |
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| 164. |
For a binary ideal lqiuid solutions,the total pressure of the solution is given asA. `P_("Total")=P_(A)^(**)+(P_(A)^(**)-P_(B)^(**))X_(A)`B. `P_("Total")=P_(B)^(**)+(P_(A)^(**)-P_(B)^(**))X_(A)`C. `P_("Total")=P_(A)^(**)+(P_(B)^(**)-P_(A)^(**))X_(A)`D. `P_("Total")=P_(B)^(**)+(P_(B)^(**)-P_(A)^(**))X_(A)` |
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Answer» Correct Answer - B Since we know `PT=P_(A)^(**)X_(A)+P_(B)^(**)X_(B)` `=P_(A)^(**)X_(A)+P_(B)^(**)(1-X_(A))` `P_(Total)=P_(B)^(**)+(P_(A)^(**)-P_(B)^(**))X_(A)` |
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| 165. |
Which of the following behaviour is true about the ideal binary liquid solution?A. Plot of `P_("Total") vs X_(A)` is non-linearB. Plot of `P_("Total") vs X_(B)` is linearC. Plot of `P_("Total")vs X_(B)` is linear with slope `= 0`D. Plot of `P_("Total") vs X_(A)` is linear with slope `= 0` |
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Answer» Correct Answer - B For a ideal binary solution `P_("Total") = p_(A)^(@) X_(A) +p_(B)^(@) X_(B)` `:.` plot of `P_("Total")` is `X_(A)` or `P_("Total")` vs `X_(B)` is a straight line |
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| 166. |
Which of the following plots represent the behaviour of an ideal binary liquid solutions?A. Plot of`P_("total")vsY_(A)`(mol fraction of`A`in vapour phase)B. Plot of`P_("total")vsY_(B)`is linearC. Plot of`(1)/(P_("total"))vsY_(A)`is linearD. Plot of`(1)/(P_("total"))vsY_(B)`is non- linear |
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Answer» Correct Answer - C `Y_(A)=(P_(A)^(@).X_(A))/(P_("total"))` Graph of `YA Vs (1)/(P_("total"))` is linear |
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| 167. |
What is molarity os the resulating solution obtained by mixing 2.5 L of 0.5 M urea solution and 500 mL of 2M urea solution ? |
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Answer» Molarity of solution on mixing : `M_(3)=(M_(1)V_(1)+M_(2)V_(2))/((V_(1)+V_(2))` According to available dete: `M_(1)=0.5M,V_(1)=2.5L,M_(2)=2M,V_(2)=500mL=0.5L` `M_(3)=((0.5Mxx2.5L)+(2Mxx0.5L))/((2.5L+0.5L))=((2.25ML))/((3.0L))=0.75M` |
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| 168. |
The mole fraction of a solute in bezene solvent is 0.2. The molality of the solution will be:A. 14B. 3.2C. 1.4D. 2 |
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Answer» Correct Answer - b `X_(B)=n_(B)/(n_(B)+n_(A))` `0.2=n_(B)/(n_(b)+1000//78)` `1/0.2=(n_(B)+12.82)/n_(B)` `5=1+(12.82)/n_(B)n_(B)=(12.82)/4=3.2` (Mass of solvent is 1000 g) |
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| 169. |
How many grams of a dibasic acid (mol. Mass `200`) should be present in `100 mL` of the aqueous solution to give `0.1 N` solution.A. 1 gB. 2 gC. 10 gD. 20 g |
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Answer» Correct Answer - A Equivalent weight of dibasic acid `=("Molecular weight")/(2)` `E=(200)/(2)=100` Stremgth `=0.1N, m = ? V = 100 mL` Normality (N) `=("Mass")/(E )xx(1000)/(V(L))` `w=(E NV)/(1000)=(100xx100xx0.1)/(1000)=1g` |
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| 170. |
How many gram of dibasic acid (mol. Mass=200) should be present in 100ml of the aqueous solution to make it 0.1 N?A. 1 gB. 2 gC. 10 gD. 20 g |
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Answer» Correct Answer - 1 |
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| 171. |
How many g of dibasic acid (mol.wt.`200)` should be present in `100ml`of its aqueous solution to give decinormal strength?A. `1g`B. `2g`C. `10g`D. `20g` |
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Answer» Correct Answer - A `N=(w)/(ExxV(l))rArr0.1=(w)/(100xx0.1)rArr w=1g` |
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| 172. |
How many kilograms of wet `NaOH` containing `12%` water are required to prepare 60 litre of 0.50 N solution ?A. `1.36 Kg`B. `1.50 Kg`C. `2.40 Kg`D. 4.16 |
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Answer» Correct Answer - A One litre if `0.50` N NaOH contains `=0.50xx40g =20g =0.020kg` `:. 60` litre of `0.50` NaOH contain `=0.020xx60 kg=1.20kg` NaOH since the given NaOH contain`12 %` water the amount of pure NAOH in `100 kg`of the given NaOH =`100-12=88 kg` Thus `88kg` of pure NaOH is present in `100kg` wet NaOH `:. 1.20 Kg` of pure NaOH is present in `(100)/(88)xx1.20=1.36` kg of wet NaOH |
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| 173. |
How many gram of NH3 (g) will be required to add in 500 ml, 0.6 N H2 SO4 to reduce the normality of H2SO4 to 0.1 N. |
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Answer» As per question the neutralised acid is equivalent to 500ml 0.5N H2SO4. 2NH3 + H2SO4,= (NH4)2SO4 By this balanced equation we get that 1mol H2SO4 requires 2 mol or 34g NH3 for nutralisation. 1000 ml 2N H2SO4 solutiin contains 1mol H2SO4. So 1000ml 2N H2SO4 requires 34g NH3 for nutralisation. 500ml 0.5N H2SO4 requires (34×500×0.5)/(1000×2) g NH3 =4.25g NH3. B |
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| 174. |
The elevation in boiling point of a solution 13.44 g of `CuCI_(2)` in 1 kg of water using the information (Molecular mass or `CuCI_(2)=13.4 "and " K_(b)=0.52" K molal "^(-01)` will beA. `0.16`B. `0.05`C. `0.1`D. `0.2` |
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Answer» Correct Answer - a `{:((a),CuCI_(2),hArr,Cu^(2+),+,2CI^(-)),("Initial",CuCI_(2),,0,,0),("molar conc",,,,,1),("Eqm.molar",,,,,),("conc",(1-alpha),,alpha,, 2alpha):}` `i=(1-alpha)+alpha+2alpha=1+2alpha` Assuming 100% dissociation `(alpha=1)` `i-1+2xx1=3` `DeltaT_(b)=ixxK_(b)xxm` `=(3xx(0.52"K kg mol"^(-1))xx(13.44g))/((134.4 "g mol"^(-1))xx(1 kg))` `=0.156 K ~~0.16 K` |
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| 175. |
The elevation in boiling point, when `13.44 g` of freshly prepared `CuCI_(2)` are added to one kilogram of water, is [Some useful data, `K_(b) (H_(2)O) = 0.52 kg K mol^(-1), "mol.wt of" CuCI_(2) = 134.4 gm]`A. `0.05`B. `0.1`C. `0.16`D. `0.21` |
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Answer» Correct Answer - C `DeltaT_(b) = iK_(b) m` `=3 x 0.52 xx (13.44//134.4)/(1)` `= 0.156 K` `=0.16K` |
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| 176. |
The elevation in boiling point, when `13.44 g` of freshly prepared `CuCI_(2)` are added to one kilogram of water, is [Some useful data, `K_(b) (H_(2)O) = 0.52 kg K mol^(-1), "mol.wt of" CuCI_(2) = 134.4 gm]`A. `0.05`B. `0.10`C. `0.16`D. `0.20` |
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Answer» Correct Answer - C `x(CuCl_(2))=(13.44)/(134.4)=0.1` `CuCl_(2)(aq)hArrCu^(2+)(aq)+2Cl^(-)(aq)` `i = 3` `DeltaT_(b)=iK_(b)xxm` `=(3xx0.52xx0.1)/(1)=0.156~~0.16` |
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| 177. |
The elevation in boiling point, when 13.44g of freshly prepared `CuCl_(2)` are added to one kilogram of water, is [Some useful data, `K_(b)(H_(2)O)=0.52"kg K mol"^(-1),"mol.wt. of CuCl"_(2)=134.4gm]`A. 0.05B. `0.10`C. 0.16D. `0.20` |
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Answer» Correct Answer - C `i = 3`, assuming complete ionization of `CuCl_(2)` `DeltaT=ixxK_(b)xx(w_(B)xx1000)/(m_(B)xxw_(A))` `=3xx0.52xx(13.44xx1000)/(134.4xx1000)=0.156~~0.16`] |
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| 178. |
1.82 g of a metal required 32.5 mL of N-HCl to dissolve it what is the equivalent weight of metal?A. 65B. 75C. 56D. 90 |
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Answer» Correct Answer - C `therefore " Meq. of metal"="Meq. of "HCL " or "1.82/Exx1000=32.5xx1` `therefore " E"=56` |
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| 179. |
Bleeding is stopped by the application of ferric chloride. This is becauseA. the blood starts flowing in opposite directionB. the blood reacts and forms a solid, which seals the blood vesselC. the blood is coagulated and thus the blood vessel is sealedD. the ferric chloride seals the blood vessel |
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Answer» Correct Answer - C Blood is a colloidal solution containing a -ve charge colloidal particle (Albuminoid), bleeding can be stopped by use alum or `FeCl_(3)` solution. The addition of `Al^(3+)` or `Fe^(3+)` causes coagulation of blood, so bleeding stops. |
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| 180. |
Which of the following ielectrolytes is least effective in causing flocculation of ferric hydroxide sol?A. `K_(4)[Fe(CN)_(6)]`B. `K_(2)CrO_(4)`C. KBrD. `K_(2)SO_(4)` |
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Answer» Correct Answer - C Ferric hydroxide sol is + vely charged sol. |
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| 181. |
Which of the following is most effective in coagulating a ferric hydroxide sol ?A. KClB. `KNO_(2)`C. `K_(2)SO_(4)`D. `K_(3)[Fe(CN)_(6)]` |
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Answer» Correct Answer - D Effectiveness of coagulation by electrolyte `prop` charge on coagulating ion. |
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| 182. |
In the layer of the atmosphere, there is a great deal of dust. When the weather is fine, it is possible to see the magnificent red colour of the setting sun. what have these observation to do with colloids? |
| Answer» Dust in the atmosphere is often colloidal. When the sun is low down on the horizon, light from it has to pass through a great deal of dust to reach your eyes. The blue part of the light is scattered away from your eyes. You see the red part of the spectrum, which remains. Red sunsets are the Tyndall effect on a large scale. | |
| 183. |
Classify the following sols according to theirs charges: (a) gold sol (b) ferric hydroxide sol (c) gelatine (d) blood (e) sulphur (f) arsenious sulphide (g) titanium oxide. |
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Answer» Negatively charged colloidal sol : (a), (c), (d), (e), (f). positive charged colloidal sol : (b), (g). |
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| 184. |
A molal solution is one that contains 1 mol of a solute inA. `1000 g` of `solvent`B. `1 L` of solventC. `1 L` of solutionD. `22.4 L` of solution |
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Answer» Correct Answer - A The number of moles of the solute present per kilogram of the solvent is known as its molality. `m=(W_("solute"))/(Mw_("solute")) xx 1/W_("solvent in kg")` =`(W_("solute"))/(Mw_("solute")) xx 1/W_("solvent in g")xx 1000` |
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| 185. |
A sample of `Na_(2)CO_(3).H_(2)O` weighing `0.62 g` is added to `100 mL` of `0.1 N H_(2)SO_(4)`. The resulting solution will beA. acidicB. basicC. amphotericD. neutral |
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Answer» Correct Answer - 4 Equivalent mass of salt `(Na_(2)^(+))_(2)CO_(3)^(2-). H_(2)O` is `("Formula mass")/("Total positive charge")` `=124 u//2=62 u` Thus, Equivalents of `Na_(2)CO_(3).H_(2)O- ("Mass")/("Gram equivalent mass")` `=(0.62 g)/(62" g "eq.^(-1))=0.01` meq of `Na_(2)CO_(3).H_(2)O=(1000) eq` `=(1000) (0.01)` `=10` meq of `H_(2)SO_(4)=V_(mL)xxN` `=(100)(0.1)` `=10` Since both acid and base have equal number of meqs, they will completely neutralize each other to form salt, `Na_(2)SO_(4)`, which does not hydrolyze. Thus, the resulting aqueous solution is neutral |
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| 186. |
A molal solution is one that contains 1 mol of a solute dissolved inA. 22.4 L of the solutionB. 1 L of the solutionC. 1 L of the solventD. 1000 g of the solvent |
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Answer» Correct Answer - 4 Molarity (m) is defined as the number of moles of the solute per kilogram (kg or 1000 g) of the solvent and is expresses as Molarity `(m)=("Moles of solute")/("Mass of solvent in kg")` For example, `1.00 m (or 1.00 mol kg^(-1))` solution of KCl means that 1 mol `(74.5 g)` of `KCl` is dissolved in every `1 kg` of water. |
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| 187. |
The latent heat of vaporisation of water is `9700 "Cal/mole"` and if the b.p.is `100^(@)C`, ebullioscopic constant of water isA. `0.513^(@)C`B. `1.026^(@)C`C. `10.26^(@)C`D. `1.832^(@)C` |
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Answer» Correct Answer - A `K_(b)=(M_(1)RT_(0)^(2))/(100DeltaH_(v))=(18xx1.987xx(373)^(2))/(1000xx9700)=0.5313^(@)C` |
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| 188. |
The osmotic pressure of a solution containing `40 g` of solute `("molecular mass 246")` per litre at `27^(@)C` is `(R=0.0822 atm L mol^(-1))`A. `3.0 atm`B. `4.0 atm`C. `2.0 atm`D. `1.0 atm` |
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Answer» Correct Answer - B `pi=CRT=((W_(2)/(Mw_(2)))RT)/V` Given`W_(2)=40 g` `Mw_(2) = 246` `T=27^(@)C = 300 K` `V= 1 L` Substituting all the values , we get `pi=(40/246 )xx 0.082 xx 300 =4 atm` |
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| 189. |
Assertion (A): The molecular mass of polymers cannot be calculated using the boiling point or freezing point method. Reason (R ): The boiling point method for determining the molecular masses is used for compounds stable at high temperature.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is correct, but `(R )` is correct. |
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Answer» Correct Answer - A Learn as a fact. The molecular mas of large molecules is determined using the osmotic pressure technique. The molecular mass of smaller molecules is determined using the freezing point lowering method. |
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| 190. |
What happents to freezing point temperature of benzene what naphthalene is added to it ?A. increases with increases in temperature.B. decreasesC. remians unchangedD. first decreases and then increases. |
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Answer» Correct Answer - b Freezing point temperatue of benzene decreases. |
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| 191. |
A 0.1539 molal aqueus solution of cane sugar `("mol mass" = 342 g mol^(-1))` has a freezing point of 271 K while freezing point ofpure water is 273.15 K. What will be the freezing point of an aqueus solution containing 5 g of glucose `("mol. Mass" =180 g mol^(-1))` per 100 g of water? |
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Answer» `Delta T_(f)=K_(f)xxm` `DeltaT_(f)=(273.1-271)=2.15 K, M=0.1539 molal` `(2.15K)=K_(f)xx0.1539 M` `m=((5g)/(180 g mol^(-1)))/((0.1 kg))=0.278m` `DeltaT_(f)=K_(f)xx0.278m` Divding eqn (ii) by eqn (i) `(DeltaT_(f))/((2.15K))=(K_(f)xx(0.278m))/(K_(f)xx(0.1539m))` `DeltaT_(f)=((0.278m)xx(2.15K))/((0.139m))=3.88 K` Freezing point temperature of the solution=(273.15-3.88)=269.27 K |
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| 192. |
Assertion (A): The increasing pressure on water decreases its freezing point. Reason (R ):The density of water is maximum at `273 K`.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is correct, but `(R )` is correct. |
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Answer» Correct Answer - C Since `P prop (1/T)`, therefore on increasing the pressure, the freezing temperature decreases. |
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| 193. |
The depression in freezing point of `0.01 m` aqueous `CH_(3)CooH` solution is `0.02046^(@)`, `1 m` urea solution freezes at `-1.86^(@)C`. Assuming molality equal to molarity, `pH` of `CH_(3)COOH` solution isA. 2B. 3C. `3.2`D. `4.2` |
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Answer» For urea `DeltaT_(t)=K_(t)xxm` or `K_(t)=(DeltaT_(t))/(m)=(1.86)/(1)=1.86` Now for `CH_(3)COOH` `DeltaT_(t)=i K_(t)m` so `i=(0.02046)/(1.86xx0.01)=1.1` Now `i=1+alpha` so `alpha=1.1-1=0.1` Now `{:(CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),(" "C,,0,,0),(C-Calpha,,Calpha,,Calpha):}` `[H^(+)]=Calpha=0.1xx0.1=0.001` so `pH=3`. |
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| 194. |
What do you mean by ideal solution? |
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Answer» Ideal solution is a solution which obeys Raoult’s law over the entire range of concentration and temperature, i.e., for an ideal solution having two volatile components A and B. PA = P°A XA , PB = P°B XB, PS = PA + PB = P°A XA+ P°B XB |
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| 195. |
(a) The mole fraction of water in a mixture containing equal number of moles of water and ethanol is (i) 1 (ii) 0.5 (iii) 2 (iv) 0.25(b) The following are the vapour pressure curves of a pure solvent and a solution of a non-volatile solute in it.Based on the above curves answer the following questions: (i) What do the curves A and B indicate? (ii) Explain why the value of TB is greater than that of Tb0. |
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Answer» (a) (ii) 0.5 (b) (i) A-Vapour pressure curve of solvent Vapour pressure curve of solution (ii) Due to the presence of a non-volatile solute vapour pressure of solution is less than solvent and the boiling point is increased. |
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| 196. |
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of 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=62g"//"mol"^(-1))`? |
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Answer» Correct Answer - A `DeltaT_(f)=6` `6=1086xx(x)/(4)` `x=12.90"mole"` Molar mass of ethylene glycol=62 Mass of ethylene glycol=800 |
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| 197. |
The melting points of most of the solid substances increases with an increase of pressure acting on them However, ice melts at a temperature lower than its usual melting point, when the pressure increases. This is Because :A. Ice is less denser than waterB. Pressure generates heatC. The bonds break undr pressureD. Ice is not a true solid |
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Answer» Correct Answer - A This is due to cage like structure ice. |
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| 198. |
What are ideal solution? Give example. |
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Answer» An ideal solution is a solution in which each component i.e., the solute as well as the solvent obeys the Raoult’s law over the entire range of concentration. |
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| 199. |
Which of the following indicators is used in the titration of `Na_(2)S_(2)O_(3)` against `CuSO_(4)` solution?A. StarchB. Barium diphenylamine sulphonateC. Cresol redD. Iodine solution |
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Answer» Correct Answer - 1 At the end point of an iodine titration with sodium thiosuphate, the iodine is all converted to iodide ions. Thus the bluish colour produced by the starch indicator with iodine disappears and the solution becomes colourless. If the indicater turns blue again. This is because some atmospheric oxidation has taken place forming `I_(2)`, which reacts with the starch to give the blue colour again. `2I^(-)+1/2 O_(2)+2H^(+) rarr I_(2)+H_(2)O` The end point of the titration is usually taken as being when the colour disappears and the solution remains colourless for half a minute. |
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| 200. |
Give the reason behind the lowering of vapour pressure in the dissolution of NaCl in water? |
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Answer» NaCI is a non volatile solute. When a non volatile solute is dissolved in pure solvent, the vapour pressure of pure solvent will decrease. In such solution, vapour pressure of the solution will depend only on the solvent molecules as the solute is non-volatile. |
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