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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.
| 751. |
The correct representation for solubility product of `SnS_(2)` isA. `[Sn^(4+)] [S^(2-)]^(2)`B. `[Sn^(2+)] [S^(2-)]^(2)`C. `[Sn^(2+)][2S^(2-)]`D. `[Sn^(4+)][2S^(2-)]^(2)` |
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Answer» Correct Answer - A `SnS_(2) hArr Sn^(4+) + 2S^(2-)` `:. K_(sp) = [Sn^(4+)] [S^(2-)]^(2)`. |
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| 752. |
What is the correct representation for the solubility product of `SnS_(2)` ?A. `[Sn^(2+)][S^(2-)]^(2)`B. `[Sn^(4+)][S^(2-)]^(2)`C. `[Sn^(2+)][2S^(2-)]`D. `[Sn^(4+)][2S^(2-)]^(2)` |
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Answer» Correct Answer - B The solubility equilibrium of `SnS_(2) ` is `SnS_(2) hArr Sn^(4+)+2S^(2-)` Solubility product `K_(sp)=[Sn^(4+)][S^(2-)]^(2)` |
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| 753. |
Conjugate acid of `SO_(4)^(2-)` isA. `HSO_(4)^(-)`B. `HSO_(4)`C. `H_(2)SO_(4)`D. `SO_(4)^(2-)` |
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Answer» Correct Answer - A Conjugate acid of `SO_(4)^(2-)` is `HSO_(4)^(-)` |
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| 754. |
If pH of a solution decreases from 5 to 2, then it isA. diluted 1000 timesB. concentrated 1000 timesC. diluted 100 timesD. concentrated 100 times |
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Answer» Correct Answer - B When `pH =5,[H^(+)]=10^(-5)` when `pH=2, [H^(+)]=10^(-2)` when the pH decreases from 5 to 2 the solution is concentrated `10^(3)(10^(-5)xx 10^(3) =10^(-2)0` times i.e., 1000 times |
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| 755. |
Calculate the equilibrium constants of each of the indicated species necessary to reduce an initial `0.2M Zn^(2+)` solution to `1.0 xx 10^(-4)Zn^(2+)`. a. `Nh_(3)` and `Zn(NH_(3))_(4)^(2+)` (assume no partial complexation) b. `overset(Theta)OH` in equilibrium with `Zn(OH)_(2)(s)`. c. `overset(Theta)OH` and `Zn(OH)_(4)^(2-)`. d. Calculate `[overset(Theta)OH]` which would be produced by each equilibrium concentration of `NH_(3)` in part (a). Predict whether `Zn(OH)_(2)` or `Zn(OH)_(4)^(2-)` would form in preference to `Zn(NH_(3))_(4)^(2+)` upon addition of suficient `NH_(3)` to produce the equilibrium concentration calculated in part(a). e. Explain what would be observeed if concentrated `NH_(3)` solution were added slowely to `0.2M` solution of `Zn^(2+)`. Given. `K_(f)Zn(NH_(3))_(4)^(2+) = 5 xx 10^(8)`. `K_(sp)ZN(OH)_(2) = 1.8 xx 10^(-14)`. `K_(f)Zn(OH)_(4)^(2-) = 5 xx 10^(14)`. `K_(b) NH_(4)OH = 1.8 xx 10^(-5)`. |
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Answer» d. `[Zn^(2+)]` final `=1 xx 10^(-4)M` `[Zn(NH_(3))_(4)^(2+)]aq = 0.2 - 10^(-4) ~~ 0.2` `Zn^(2+) + 4NH_(3) hArr Zn(NH_(3))_(4)^(2+)` `K_(f) = ([Zn(NH_(3))_(4)^(2+)])/([Zn^(2+)][NH_(3)]^(4))` `5 xx 10^(8) =(0.2)/((10^(-4))[NH_(3)]^(4))` `(NH_(3))^(4) = 4 xx 10^(-6) = 400 xx 10^(-8)` `(NH_(3)) = 4.5 xx 10^(-2)M` b. `Zn(OH)_(2) rarr Zn^(2+) +2 overset(Theta)OH` `K_(sp) = (1.8 xx 10^(-14))/(10^(-4)) = 1.8 xx 10^(-10)` `[overset(Theta)OH] = 1.3 xx 10^(-5)M` c. `Zn^(2+) + 4overset(Theta)OH rarr Zn(OH)_(4)^(2-)` `K_(f) - ([Zn(OH)_(4)^(2-)])/([Zn^(2+)][overset(Theta)OH]^(4))` `5 xx 10^(4) = (0.2)/((10^(-4))[overset(Theta)OH]^(4))` `rArr [overset(Theta)OH]^(4) = 4 xx 10^(-12), [overset(Theta(O)H] = 1.4 xx 10^(-3)` d. The concentration of `overset(Theta)OH` in equilibrium with the `NH_(3)` concentration of part (a) is given by, `NH_(3) + H_(2)O hArr overset(o+)NH_(4) + overset(Theta)OH`, (Let `x = [overset(o+)NH_(4)] = [overset(Theta)OH]` `K_(b) = ([overset(o+)NH_(4)][overset(Theta)OH])/([NH_(3)])` `1.8 xx 10^(-5) = (x^(2))/(4.5 xx 10^(-2))` `rArr x^(2) = 8.1 xx 10^(-7) rArr x = [overset(Theta)OH] = 9.0 xx 10^(-4)M`. `[overset(Theta)OH] (9.0 xx 10^(-4)M)` from the `4.5 xx 10^(-2)M NH_(3)` is sufficient precipitate `Zn(OH)_(2)` [part (b)] but not sufficient to from `Zn(OH_(4)^(2-))` [part (c)] e. On addition of `Nh_(3)` solution to `Zn^(2+)` solution, a precipitate of `Zn(OH)_(2)` will form, which will later dissolve to yield a clear solution containing `[Zn(NH_(3))_(4)]^(2+)` |
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| 756. |
What is the `pH` of a solution containing `0.01 mol HCI L^(-1)`? |
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Answer» `HCI` is strong acid and it is completely ionised `(100%` ionsiation) `HCI +H_(2)O rarr H_(3)O^(o+) +CI^(Θ)` `[H^(o+)] = (0.01)/(1L) = 10^(-2) M :. pH = 2.0` |
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| 757. |
Given a 0.1 M solution of each of the following. Which solution has the lowest pHA. `NaHSO_(4)`B. `NH_(4)Cl`C. HClD. `NH_(3)` |
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Answer» Correct Answer - C Because it is a strong acid. `H^(+) = 10^(-1)`. `pH = -log [H^(+)] = -log[10^(-1)] , pH = 1`. |
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| 758. |
which of the following dissolves in water to give a neutral solutionA. `(NH_(4))_(2)SO_(4)`B. `Ba(NO_(3))_(2)`C. `CrCl_(3)`D. `CuSO_(4)` |
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Answer» Correct Answer - B `Ba(NO_(3))_(2)` does not undergo hydrolysis. |
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| 759. |
A 1 M aqueous solution of which molecule has the lowest pH ?A. HOClB. `H_2SO_3`C. `H_3PO_4`D. `H_2SO_4` |
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Answer» Correct Answer - d |
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| 760. |
The addition of 0.01 mol of which molecule has the lowest pH ?A. `NH_3`B. `HONH_2`C. `CH_3NH_2`D. `H_2N NH_2` |
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Answer» Correct Answer - c |
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| 761. |
`H^(+)` is aA. Lewis acidB. Lewis baseC. Bronsted-Lowry baseD. None of the above |
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Answer» Correct Answer - A Because it is a electron pair acceptor. |
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| 762. |
For two acids A and B, `pK_(a) = 1.2, pK_(b) = 2.8` respectively in value , then which is trueA. A and B both are equally acidicB. A is strongly than AC. B is stronger than AD. Neither A nor B is strong |
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Answer» Correct Answer - B The value of `pK_(a)` for strong acid is less. |
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| 763. |
Lewis base isA. `CO_(2)`B. `SO_(3)`C. `SO_(2)`D. ROH |
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Answer» Correct Answer - D ROH is a Lewis base because it has an lone pair of electron. |
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| 764. |
For which equation is the equilibrium constant equal to `K_a` for the ammonium ion, `NH_4^+`?A. `NH_4^+(aq)+OH^(-)(aq)hArr NH_3(aq)+H_2O(l)`B. `NH_4^+(aq)+H_2O(l)hArr NH_3(aq)+H_3O^+(aq)`C. `NH_3(aq)+H_2O(l)hArr NH_4^+(aq)+OH^(-)(aq)`D. `NH_3(aq)+H_3O^+(aq)hArr NH_4^+(aq)+H_2O(l)` |
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Answer» Correct Answer - b |
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| 765. |
Which of the following will give maximum ions in solution ?A. `KI_(3)`B. `CuSO_(4)`C. `FeCl_(3)`D. `K_(2)HgI_(4)` |
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Answer» Correct Answer - C `(A)" "KI_(3) hArr K^(+)+I_(3)^(-)" "`(Two ions) `(B) CuSO_(4)hArr Cu^(2+)+SO_(4)^(2-)" "`(Two ions ) `(C ) FeCl_(3) hArr Fe^(3+)+Cl^(-) " "`(Four ions ) `(D) K_2)[HgI_(4)] hArr 2K^(+)+[HgI_(4)]^(-)" "` (Three ions ) |
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| 766. |
The suitable for strong acid and weak base isA. Methyl orangeB. Methyl redC. Phenol redD. Phenolphthalein |
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Answer» Correct Answer - A::B Solution become acidic and methyl orange act on acidic pH. Methyl red can also be used. |
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| 767. |
Which is not a electrophileA. `AlCl_(3)`B. `BF_(3)`C. `(CH_(3))_(3)C^(+)`D. `NH_(3)` |
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Answer» Correct Answer - D Electron donating species called nucleophile. `NH_(3)` have a lone pair of electron. |
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| 768. |
`pH` of a mixture containing `0.10 M X^(-)` and `0.20 M HX` is: `[pK_(b)(X^(-))=4]`A. 4+ log 2B. 4- log 2C. 10 + log 2D. 10 - log 2 |
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Answer» Correct Answer - d |
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| 769. |
Which of the following metal sulphides has maximum solubility in water?A. `CdS(K_(sp)=36xx10^(-30))`B. `FeS(K_(sp)=11xx10^(20))`C. `HgS(K_(sp)=32xx10^(-54))`D. `ZnS(K_(sp)=11xx10^(-22))` |
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Answer» Correct Answer - B The salt with highest `K_(sp)` is most soluble. Thus, FeS will be most soluble. |
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| 770. |
`H_(3)BO_(3)` is :A. monobasic and weak Lewis acidB. monobasic and weak Bronsted acidC. monobasic and strong Lewis acidD. tribasic and weak Bronsted acid. |
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Answer» Correct Answer - A `H_(3)BO_(3)` in watter gives `H^(+)` ions as follows `H_(3)BO_(3)+H_(2)O rarr [B(OH)_(4)]^(-)+H^(+)` It is monobasic as one molecule of the acid gives one `H^(+)` ion. As `H_(3)BO_(3)` or `B(OH)_(3)` takes up a pair of electrons from `OH^(-)` , it is a Lewis acid. |
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| 771. |
`H_(3)BO_(3)` in water acts as aA. Lewis acidB. Bronsted acidC. Arrhenius acidD. Both (B) and (C ) |
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Answer» Correct Answer - A `B(OH)_(3) ` or `H_(3)BO_(3)+H_(2)O hArr [B(OH)_(4)]^(-)+H^(+)` Thus, `H_(3)BO_(3)` acts only as a Lewis acid. It accepts a pair of electrons from `OH^(-)` |
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| 772. |
The `pH` of `0.1 M` solution of the following salts increases in the orderA. `NaCl lt NH_(4)Cl lt NaCN lt HCl`B. `HClltNH_(4)ClltNaClltNaCN`C. `NaCNltNH_(4)ClltNaClltHCl`D. `HClltNaClltNaCNltNH_(4)Cl` |
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Answer» Correct Answer - B `NH_(4)Cl` undergoes cationic hydrolysis hence pH is `lt7` because the solution due to cationic hydrolysis is acidic NaCN undergoes anionic and NaCl is neutral solution. Hence the pH of given solutions will increases as `HCl lt NH_(4)Cl ltNaCl lt NaCN` |
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| 773. |
`pH` of a mixture containing `0.10 M X^(-)` and `0.20 M HX` is: `[pK_(b)(X^(-))=4]`A. `4+log2`B. `4-log2`C. `10+log2`D. `10-log2` |
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Answer» `pOH=pK_(b)+log((["Salt"])/(["Base"]))` Let a mol litre^(-1) be concentration of salt., then concentration of base `=(0.29-a)mol//L`. `4.4=-log1.8xx10^(-5)+log((a)/((0.29-a)))` `:.a=0.09` `["Salt"]=0.9M` & `["Base"]=0.29-0.09=0.20M` |
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| 774. |
The solubility of `A_(2)X_(2)` is `x` mole `dm^(-3)`. Its solubility product isA. `6y^(4)`B. `64y^(4)`C. `36y^(5)`D. `108 y ^(5)` |
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Answer» Correct Answer - D `underset("Eqm.")(A_(2)X_(3)) hArr underset(2y)(2A^(3+))+underset(3y)(3X^(2-))` `:. K_(sp)=(2y)^(2)(3y)^(3)=108 y^(5)` |
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| 775. |
Which of the following metal sulphides has maximum solubility in water?A. `CdS (K_(sp) = 36 xx 10^(-30))`B. `FeS (K_(sp) = 11 xx 10^(-20))`C. `HgS(K_(sp) = 32 xx 10^(-54))`D. `ZnS(K_(sp) = 11 xx 10^(-22))` |
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Answer» Correct Answer - B Higher the `K_(sp)`, more soluble is that compound in `H_(2)O` `:. K_(sp)` of `FeS (11 xx 10^(-20))` is highest. So, it is more soluble and has maximum solubility in `H_(2)O`. |
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| 776. |
A buffer solution made up of `BOH` of total maturity `0.29M` has `pH=9.6` and `K_(b)=1.8xx10^(-5)`.Concentration of salt and base respectively is :A. `0.09M and 0.2M`B. `0.2 M and 0.09M`C. `0.1 M and `0.19m`D. `0.19M and 0.1M` |
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Answer» `pOH=-logK_(b)+log((["Salt"])/(["Base"]))` Let a mol litre^(-1) be concentratoion of base =`(0.29-a)mol//L` `4.4=-log1.8xx10^(-5)+log((a)/((0.29-a)))` `:.a=0.09` `["Salt"]=0.9M` & `[Base]=0.29-0.09=0.20M` |
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| 777. |
The solubility of `A_(2)X_(3)` is y mol `dm^(-3)`. Its solubility product isA. `6 y^(4)`B. `64 y^(4)`C. `36 y^(5)`D. `108 y^(5)` |
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Answer» Correct Answer - D `{:(A_(2)X_(3)hArr,2A^(3+)+,3X^(2-),,K_(sp) = [A^(3+)][X^(2-)] = (2y)^(2)(3y)^(3)),(,2y,3y,):}` or `K_(sp) = 108 y^(5)` |
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| 778. |
Incorrect formula for Hydronium ion is :A. `H_3O^(+)`B. `H_9O_4^(+)`C. `H_5O_2^(+)`D. `H_4O_2^(+)` |
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Answer» Correct Answer - d |
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| 779. |
If the `K_(b)` value in the hydrolysis reaction `B^(+) + H_(2)O hArr BOH + H^(+)` is `1.0 xx 10^(-6)`, then the hydrolysis constant of the salt would beA. `1.0 xx 10^(-6)`B. `1.0 xx 10^(-7)`C. `1.0 xx 10^(-8)`D. `1.0 xx 10^(-9)` |
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Answer» Correct Answer - C For hydrolysis of `B^(+), K_(H) = (K_(w))/(K_(b)) = (10^(-14))/(10^(-6)) = 10^(-8)`. |
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| 780. |
The `pH` of `0.1 M` solution of the following salts increases in the orderA. `NaCIltNH_(4)CIltNaCNltHCI`B. `HCIltNH_(4)CIltNaCIltNaCN`C. `NaCNltNH_(4)CIltNaCIltHCI`D. `HCIltNaCIltNaCNltNH_(4)CI` |
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Answer» Correct Answer - B Higher is pH, lesser is acidic nature. Also `NH_(4)CI(aq.)` is acidic and `NaCI(aq.)` is alkaline `NaCI(aq.)` is neutral. |
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| 781. |
The set with correct order of acidity isA. `HClO lt HClO_(2) lt HClO_(3) lt HClO_(4)`B. `HClO_(4) lt HClO_(3) lt HClO_(2) lt HClO`C. `HClO lt HClO_(4) lt HClO_(3) lt HClO_(2)`D. `HClO_(4) lt HClO_(2) lt HClO_(3) lt HClO` |
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Answer» Correct Answer - A For oxyacids containing similar central atom, the acid strength increases with the increase in the number of oxygen attached to the central atom and not attached to any other atom. Alternatively, higher the oxidation number of the central atom, higher is the acidity of the species. Thus acidity follows the order `{:("Oxi. No. of Cl",HClO ,lt,HClO_(2),lt, HClO_(2), lt, HClO_(4)),(,+1,,+3,,+5,,+7):}` |
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| 782. |
The phenomenon of interaction of anions and cations furnished by a electrolyte with the `H^(+)` and `OH^(-)` ions of water to produce acidity or alkalinity is known as hydrolysis. In hydrolysis :A. the pH may either increases or decreaseB. all the salts (except those made up with strong anion and cation) undergo hydrolysisC. the variation of pH depends upon the nature of salt as well on the temperaratureD. none of these |
| Answer» Correct Answer - A::B::C | |
| 783. |
`100mL` of a clear saturated solution of `Ag_(2)SO_(4)` is added to `250mL` of a clear saturated solution of `PbCrO_(4)` .will any precipitate form and if so what ?Given `K_(sp)` values of `Ag_(2)SO_(4),Ag_(2)CrO_(4),PbCrO_(4)& PbSO_(4)` are `1.4xx10^(-5),2.4xx10^(-12),2.8xx10^(-13)` and `1.6xx10^(-8)` respectively. |
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Answer» For `Ag_(2)SO_(4)hArr underset(2s)2Ag^(+)+underset(s)SO_(4)^(2-)` `K_(sp)=4s^(3)or s=3sqrt((K_(sp))/(4))=sqrt((1.4xx10^(-5))/(4))=1.52xx10^(-2)M` for `PbCrO_(4)hArr Pb^(2+)+CrO_(4)^(2-)` `K_(sp)=s_(1)^(2) or s_(1)sqrt(K_(sp))=sqrt(2.85xx10^(-13))=5.29x10^(-7)M` In solution concentration of each ion can be given as: Thus `[Ag^(+)]=(2sxx100)/(350)=(2xx1.52xx10^(-2)xx100)/(350)=0.869xx10^(-2)M` `[SO_(4)^(2-)]=(sxx100)/(350)=(1.52xx10^(-2)xx100)/(350)=0.43xx10^(-2)M` `[Pb^(2+)]=(s_(1)xx250)/(350)=(5.29xx10^(-7)xx250)/(350)=3.78xx10^(-7)M` `[CrO_(4)^(2-)]=(s_(1)xx250)/(350)=(5.29xx10^(-7)xx250)/(350)=3.78xx10^(-7)M` It is thus evidence that, `[Ag^(+)]^(2)[CrO_(4)^(2-)]=(0.869xx10^(-2))^(2)xx(3.78xx10^(-7))` `=2.85xx10^(-11)(gtK_(sp)Ag_(2)CrO_(4))` Thus `Ag_(2)CrO_(4)` will percipitate. |
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| 784. |
Strong acids are generally used as standard solution in acid-base titrations because:A. The pH at equivalence point will be 7B. They titrate both strong and weak baseC. They form more stable solutions than weak acidsD. The salts of strong acids do not hydrolyse |
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Answer» Correct Answer - b |
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| 785. |
The best explanation for the solubility of MnS in dil. HCI is that:A. Solubility product of `MnCl_(2)` is less than that of MnS.B. Concentration of `Mn^(2+)` is lowered by the formation of complex ions with chloride ions.C. Concentration of sulphide ions is lowered by oxidation to free sulphur.D. Concentration of sulphide ions is lowered by formation of week acid `H_(2)S` |
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Answer» Correct Answer - d |
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| 786. |
Which of the following is the buffer solution ?A. `CH_(3)COOH+CH_(3)COONa`B. `CH_(3)COOH + CH_(3)COONH_(4)`C. `CH_(3)COOH +NH_(4)Cl`D. `NaOH + NaCl` |
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Answer» Correct Answer - A At weak acid and its salt with a strong acid ( or a weak base and its salt with a strong acid ) can act as a buffer solution. |
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| 787. |
`DeltaH_("fus")` is the latent heat of melting and `T_("fus")` is the melting point, then entropy of melting isA. `DeltaH_("fus")//T_("fus")`B. `DeltaH_("fus")xxT_("fus")`C. `T_("fus")//DeltaH_("fus")`D. `DeltaH_("fus")-T_("fus")` |
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Answer» Correct Answer - A `Delta S=(q)/(T)=(DeltaH_("fus"))/(T_("fus"))` |
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| 788. |
Which of the following is not a Lewis acid ?A. `AlCl_(3).6H_(2)O`B. `AlCl_(3)`C. `SnCl_(4)`D. `FeCl_(3)` |
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Answer» Correct Answer - A `AlCl_(3), SnCl_(4)` and `FeCl_(3)`act as Lewis acids because central atom in them have one vacant 3p orbital and five 3d orbitals. But in `AlCl_(3).6H_(2)O`, all these empty orbitals get occupied. Hence it cannot act as Lewis acid. |
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| 789. |
Which of the following is a state function ?A. Temperature of an ice cubeB. The amount of work in expansionC. Both (A) and (B)D. None of the above |
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Answer» Correct Answer - A Temperature is a state function but work is not a state function. |
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| 790. |
For an adiabatic processA. `q= + W`B. `Delta E ltW`C. `Delta E ge W`D. `q=0` |
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Answer» Correct Answer - D See Comprehensive Review. |
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| 791. |
Pressure is doubled in each of the following equilibrium . In which case yield is maximum ?A. `C(s)+H_(2)O(g) hArr CO(g)+H_(2)(g)`B. `2H_(2)(g)+O_(2)(g) hArr 2H_(2)O(g)`C. `2Fe(s)+3H_(2)O(g) hArr Fe_(2)O_(3)(s)+3H_(2)(g)`D. `N_(2)O_(4)(g) hArr 2NO_(2)(g)` |
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Answer» Correct Answer - B On increasing pressure, the position of equilibrium in a reversible reaction which is accompanied by a decrease in number of moles of gaseous constituents will shift in the forward direction. |
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| 792. |
In which of the reactions, the enthalpy is the leastA. `CH_(3)COOH + NaOH rArr CH_(3)COONa + H_(2)O`B. `HCl + NH_(4)OH rArr NH_(4)Cl + H_(2)O`C. `HCl + NaOH rArr NaCl + H_(2)O`D. `HCN + NH_(4)OH rarr NH_(4)CN + H_(2)O` |
| Answer» Correct Answer - C | |
| 793. |
pH of a solution produced when an aqueous soution of `pH =6` is mixed with an equal volume of an aqueous solution of `pH =3` is about `:`A. 3.3B. 4.3C. `4.0`D. 4.5 |
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Answer» Correct Answer - A Assume 1 L of each solution `[H_(3)O^(+)]` in solution of `pH =6` is `10^(-6)M` `[H_(3)O^(+)]` in solution of `pH=3` is `10^(-3) M` Total `[H_(3)O^(+)]=((10^(-3)+10^(-6))/(2))` `=5.005xx10^(-4)` `:. pH =4- lo g 5.005 ~~3.3` |
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| 794. |
pH of a solution produced when an aqueous solution of pH 6 is mixed with an equal volume of an aqueous solution of pH 3 is aboutA. 3.3B. 4.3C. `4.0`D. 4.5 |
| Answer» Correct Answer - A | |
| 795. |
The charge balance equation of species in 0.100 M acetic solution is given byA. `[H^(+)] = [OH^(-)]`B. `[H^(+)] = [CH_(3)COO^(-)]`C. `[H^(+)] = [OH^(-)] + [CH_(3)COO^(-)]`D. `2[H^(+)] = [OH^(-)] + [CH_(3)COO^(-)]` |
| Answer» Correct Answer - D | |
| 796. |
The relative basic character of the following isA. `CIO^(-) lt CIO_(2)^(-) lt CIO_(3)^(-) lt CIO_(4)^(-)`B. `CIO_(4)^(-) lt CIO_(3)^(-) lt CIO_(2)^(-) lt CIO^(-)`C. `CIO_(3)^(-) lt CIO_(4)^(-) lt CIO_(2)^(-) lt CIO^(-)`D. `CIO_(2)^(-) lt CIO^(-) lt CIO_(3)^(-) lt CIO_(4)^(-)` |
| Answer» Correct Answer - B | |
| 797. |
pH of a centimolar solution of a mono basic acid is `6`. The dissociation constant is approximately equal toA. `10^(-12)`B. `10^(-8)`C. `10^(-10)`D. `10^(-6)` |
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Answer» Correct Answer - C `P^(H) = 6 , C=0.01` `[H^(+)] = 10^(-6)` `[H^(+)] = sqrt(K_(a) xx C)` `K_(a) = 10^(-10)` |
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| 798. |
Assertion (A): A certain recation is catalysed by acids and the catalytic activity of `0.1M` solutions of the acids in water decrease in the order, `HCI, HCOOH`, and `CH_(3)COOH`. The same reaction takes place in anhydrous `NH_(3)`, but the three acids have same catalytic effect in `0.1M` solution. Reason (R) : The order of catalytic activity in water is the same as the order of acidity. in anhyrous `NH_(3)`, all the three acids are strong.A. If both (A) and (R) are correc, 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 incorrect, but (R) is correct. |
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Answer» Correct Answer - A Both `(A)` and `(R)` are correct and `(R)` is correct explanation of `(A)`. |
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| 799. |
The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferential precipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing `Cl^(-), Br^(-)` and `I^(-)` ions, if `Ag^(+)` ions are added, then out of three , the less soluble salt is precipitated first . If the addition of `Ag^(+)` ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum `K_(sp)` ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing `Cl^(-)` and `CrO_(4)^(2-)`. In order to predict which ion will precipitate first, we have to calculate the amount of `Ag^(+)` ions needed to start precipitation through the solubility product data given. When `AgNO_93)` is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding `[Ag^(+)]` to start the precipitation increases. Its concentration eventually becomes equal to the value required for `CrO_(4)^(2-)` . At this stage, practically almost the whole of `Cl^(-)` ions get precipitated . Addition of more `AgNO_(3)` causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of `Cl^(-),Br^(-)` and `I^(-)` . Further he adds gradually solid `AgNO_(3)` to this solution. He assumes that volume of the solution does not change after the addition of solid `AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2)` and `K_(sp) AgI=10^(-12)M^(2))` Experiment -2. He prepares a solution of cations `Cd^(2+)(0.2 M)` and `Bi^(3+) (0.3 m)` . Now he adds `S^(2-)` ions into the solution of cations in order to separate them by selective precipitation `Cd^(2+) `forms yellow precipitate of CdS and `Bi^(3+)` forms black precipitate of `Bi_(2)S_(3)` with `S^(2-)` ions respectively. `(K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5)` and `K_(sp)CdS =2 xx 10^(-20)M^(2))`.Answer the following questions ont he basis of the above write up. Which of the following is the correct order of appearance of coloour of the precipitate in experiment ?A. yellow `gt` pale yellow `gt` curdy whiteB. yellow `lt` pale yellow `lt` curdy whiteC. pale yellow `gt` yellow `lt` curdy whiteD. pale yellow `gt` yellow `gt` curdy white |
| Answer» Correct Answer - A | |
| 800. |
Calculate the percent error in the `[H_(3)O^(o+)]` made by neglecting the ionisation of water in `10^(-6)M NaOH` solution. |
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Answer» a. Neglecting the ionisation of `H_(@)O, [oversert(Θ)OH]` from `10^(-6)M NaOh`, and `[H_(3)O^(o+)] = 10^(-8)M`. ltbRgt b. Including the ionisation of water `x = [H_(3)O^(o+)]` and `(x+10^(-6)) = [overset(Θ)OH]` `:. (x+10^(-6)) (x) = 10^(-14)` or `x^(2) +10^(-6) x- 10^(-14) = 0` `:.x=(-10^(-6)+sqrt(10^(-12)+4xx10^(-14)))/(2)=9.9xx10^(-9)` % error `= ((10^(-8))-(9.9 xx 10^(-9)))/(9.9xx10^(-9))xx100` `= ((10xx10^(-9))-(9.9xx10^(-9)))/(9.9xx10^(-9))xx100% =1%` |
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