

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. |
In qualitative analysis, on what basis cations are grouped? |
Answer» Cations are grouped on the basis of their solubility product (Ksp). |
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752. |
Under what conditions precipitation takes place in ionic reactions? |
Answer» When ionic product is greater than solubility product. |
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753. |
Derive ionic product of water. Also find the value at 25°C |
Answer» When high current is passed through water it under goes partial dissociation. H2O ⇌ H+ + OH- Applying law of mass action, K = \(\frac{[H^+][OH^-]}{[H_2O]}\); K[H2O] = [H+][OH-] But k[H2O] = Kw ∴ Kw = [H+][OH-] Where, Kw is ionic product of water. Value of Kw at 25°C: It is found that at 25°C [H+] = [OH–] = 10-7 mol/dm3 ∴ Kw = [H+][OH–] = 10-7 × 10-7 = 10-14 (mol/dm3)2 At 25°C the value of Kw is 10-14(tmol/dm3)2 |
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754. |
For the chemical reaction `3X(g)+Y(g) hArr X_(3)Y(g)`, the amount of `X_(3)Y` at equilibrium is affected byA. temperature and pressureB. pressure onlyC. temperature onlyD. temperature pressure and catalyst. |
Answer» Correct Answer - A is the correct answer |
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755. |
How many moles of AgBr `(K_(sp)=5xx10^(-13)"mol"^(2)L^(-2))` will dissolve in 0.01 M NaBr solution ? |
Answer» Correct Answer - `5xx10^(-11) "mol" L^(-1)` Suppose solubility of AgBr in 0.01 M NaBr = s mol `L^(-1)`. Then as `AgBr rarr Ag^(+) + Br^(-)`, `[Ag^(+)]=s "mol " L^(-1) and "Tota" [Br^(-)]=0.01 + s ~~ 0.01 M` `K_(sp) = [ Ag^(+)][Br^(-)], i.e., 5xx10^(-13) = s xx0.01 or s = 5 xx 10^(-11) "mol" L^(-1)` |
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756. |
What is the range of a pH indicator in terms of its dissociation constant (Kin)? |
Answer» pH = pKin ± 1. |
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757. |
The equilibrium constant of a reaction at 27o c is 1.6 and at 127o c is 7.6 respectively. Is the reaction exothermic or endothermic? |
Answer» As the value of equilibrium constant increases with increase in temperature so, the reaction is endothermic in nature. |
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758. |
How do you explain detection of H group basic radicals based on application of solubility product. |
Answer» Group reagent: HCl → H+ + Cl– ; H2S ⇌ 2H+ + S2- Radicals: Cu2+, Bi2+, Sn2+ Here HCl is a strong electrolyte under goes complete dissociation where as H2S is a weak electrolyte under goes partial dissociation and attains equilibrium between it’s ions and molecules. In this solution, because of common ion (H+) effect supression of degree of dissociation of H2S takes place. As a result [S2-] decreases. Hence the metal radical forms it’s respective sulphides [CuS, SnS, BiS] whose solubility product is less than ionic product. Hence forms precipitate. |
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759. |
Explain detection of IV group basic radicals using application of solubility product. |
Answer» Group reagent: NH4Cl → NH4+ + Cl– ; NH4OH ⇌ NH4+ + OH– H2S ⇌ 2H+ + S2- ; H+ + OH– → H2O Radicals: Zn2+, Mn2+ Here, NH4Cl is a strong electrolyte undergoes complete dissociation, where as NH4OH– and H2S are weak electrolytes undergoes partial dissociation and attains equilibrium between their ions and molecules. Along with common ion (NH4+) effect, the H+ and OH- ion combines to form water molecules. At the same time the metals forms their respective sulphides [ZnS, MnS] Whose solubility product is less than ionic product. Hence, forms precipitate. |
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760. |
How do you explain the detection of HI group basic radicals based on application of solubility product? |
Answer» Group reagent: NH4OH ⇌ NH4+ + OH– ; NH4Cl ⇌ NH4+ + Cl– Radicals: Al3+, Cr3+, Fe3+ Here NH4Cl is a strong electrolyte under goes complete dissociation where as NH4OH is a weak electrolyte under goes partial dissociation and attains equilibrium between it’s ions and molecules. In this solution because of common ion (NH4+) effect supression of degree of dissociation of NH4OH takes place. As a result [OH–] decreases. Hence the metal radical forms it’s respective hydroxides [Al(OH)3, Cr(OH)3, Fe(OH)3]. Whose solubility product is less than ionic product. Hence forms precipitate. |
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761. |
What is the effect of temperature on the reactions?N2(g) + O2(g) ⇌ 2NO3(g) - Heat |
Answer» The increase in temperature favours forward reaction because the reaction is endothermic. |
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762. |
What happens to the solubility of AgCl in water if NaCI solution is added to it? |
Answer» Solubility of AgCI decreases due to common ion effect. |
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763. |
How do you explain detection of 1 group basic radicals based on application and solubility product. |
Answer» Group reagent: Dil. HCl → H+ + Cl– Redicals present: Pb2+, Hg2+, Cd2+ The group reagent HCl is a strong electrolyte under goes complete dissociation. The metal radicals forms it’s respective chlorides [PbCl2, HgCl2, CdCl2]. Whose solubility product is less than ionic product. Hence, forms precipitate. |
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764. |
In presence of HCl, `H_(2)S` results the precipitation of group-2 radicals but not group-4 radicals during qualitative analysis. It is due toA. higher concentration of `H^(+)`B. lower concentration of `H^(+)`C. higher concentration of `S^(2-)`D. lower concentration of `S^(2-)` |
Answer» Correct Answer - D Dissociation of `H_(2)S` is suppressed in presence of HCl due to common ion effect. This decreases the `S^(2-)` ion concentration and hence only group 2 radicals are precipitated as they have low solubility products. |
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765. |
Why in Group V of qualitative analysis, sufficient `NH_(4)OH` solution should be added before adding `(NH_(4))_(2)CO_(3)` solution ? |
Answer» This is done to convert `NH_(4)HCO_(3)` usually present in large amounts along with `(NH_(4))_(2)CO_(3) ` to `(NH_(4))_(2)CO_(3)` `NH_(4)HCO_(3)+NH_(4)OH rarr (NH_(4))_(2)CO_(3)+H_(2)O` |
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766. |
What happens to the solubility to `CO_(2)` in water if temperature is increased ? |
Answer» Correct Answer - It decreases | |
767. |
For an exothermic reaction, what happens to the equilibrium constant if temperature is increased. |
Answer» \(\because\) K = \(\frac{K_f}{K_b}\) In exothermic reaction, with the increase in temperature, equilibrium constant for backward reaction (Kb) increases much more than Kf. Thus, equilibrium constant (K) decreases with increase in temperature for an exothermic reaction. |
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768. |
What qualitative information can you obtain from the magnitude of equilibrium constant? |
Answer» (i) Large value of equilibrium constant (>103) shows that forward reaction is favoured, i.e., concentration of products is much larger than that of the reactants at equilibrium. (ii) Intermediate value of K(10-3 to 103) shows that the concentration of the reactants and products are comparable, (iii) Low value of K(<10-3)shows that backward reaction- is favoured, i.e, concentration of reactants is much larger than that of the products. |
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769. |
Define degree of ionization or dissociation. |
Answer» It is the ratio of number of ions or molecules produced to the total number of molecules. |
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770. |
What happens to the process if it subjected to a change given in the brackets? Dissolution of NaOH in water (Temperature is increased) |
Answer» Solubility will decrease because it is an exothermic process. |
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771. |
Some process are given below. What happens to the process if it is subjected to a change given in the barckets ? (ii) Dissolution of ` Ice overset (M.pt) hArr " Water ( Pressure is increased ) "` (ii) Dissolution of NaOH in water ( Temperature is increased) (iii) `N_(2) (g) + O_(2) (g) hArr 2 NO (g) - 180* 7 kJ ` (pressure is increased and temperature is decreased). |
Answer» (i) Equilibrium will shift in the forward direction , i.e., more of ice will melt. (ii) Solubility will decrease because it is an exothermic process. (iii) Pressurehas no effect. Decrease of temperature will shift the equilibrium in the backward direction. |
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772. |
What happens to the process if it subjected to a change given in the brackets? N2(g) + O2(g) ⇌ 2NO(g) -180.7kJ(Pressure is increased and temperature is decreased) |
Answer» Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction. |
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773. |
What is difference between strong electrolyte, and weak electrolyte? |
Answer» Strong electrolytes can dissociate into ions completely in aqueous solution whereas weak electrolytes do not dissociate into ions completely and there exist an equilibrium between ioins and unionized molecules, called Ionic Equilibrium. |
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774. |
Give an example of sparingly soluble salt? |
Answer» AgCl.......... | |
775. |
If the solubility product of silver chloride is 1.8 × 10-10. What is the solubility of silver ion if concentration of Cl– is 0.01 molar. |
Answer» Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+]0.01 [Ag+] = \(\frac{1.8\times 10^{-10}}{0.01}\) = 180 x 10-10 = 1.8 x 10-8 mol/dm3 |
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776. |
What happens to the process if it subjected to a change given in the brackets?Ice \(\overset{M.pt}{\rightleftharpoons}\) Water (Pressure is increased) |
Answer» Equilibrium will shift in the forward direction, i.e., more of ice will melt. |
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777. |
Give any two differences between strong and weak electrolyte. |
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778. |
The solubility of AgCl in water at 250 C is found to be 1.06*10-5 moles per litre. Calculate the solubility product of AgCl at this temperature. |
Answer» AgCl ionizes completely in the solution as AgCl ----> Ag+ +Cl— One mole of AgCl in the solution gives 1 mole of Ag+ ions and 1 mole of Cl—ions. Now as the solubility of AgCl= 1.06*10—5 moles per litre. Therefore [Ag+]=1.06*10—5 moles per litre and [Cl--] = 1.06*10—5 moles per litre. Ksp for AgCl=[Ag+] [Cl--]=1.06*10—5 *1.06*10—5 =1.1*10—10 |
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779. |
What is common ion effect and how it can be used to purifies impure sodium chloride? |
Answer» To an ionic equilibrium, AB ----> A+ +B+,a salt containing a common ion ( AC or BD is added, the equilibrium shifts in the backward direction. This is called common ion effect. On passing hydrogen chloride gas through a saturated solution o sodium chloride, the concentration of chloride ions will be increased. This increases the ionic product of NaCl and the solid salt will be precipitated. |
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780. |
What is common ion effect ? Give example. |
Answer» Supression in degree of dissociation of a weak electrolyte by addition of a common ion is called common ion effect. Example:- CH3COOH and CH3COONa |
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781. |
What is common ion effect? |
Answer» Shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium. |
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782. |
Acids are useful on one hand but sometimes they are troubleshooters on the other hand. Therefore, in a number of cases, acids have to be neutralized with the help of suitable bases. In nature, a number of plants grow some of which contain acids while some other contain bases. Now answer the following questions: (i) Which acid is called stomach acid? What is its use in the body and how does it harm our system? What is the treatment taken? (ii) What is acid rain? How does it affect aquatic life and soil? What treatment has to be given to the soil? (iii) Why do we get pain when we are stung by a honey bee or an ant? What is the treatment given? (iv) If you go on hiking and trekking and by chance, you touch nettle plant whose leaves have stinging hair, you get pain. Why? What treatment will you take there? |
Answer» (i) HCl. It helps in digestion of food. Sometimes stomach produces too much acid and this causes pain and irritation. To neutralize the acidity antacids such as magnesium hydroxide are taken. (ii) When acidic gases present in the air dissolve in the rain water, such that the pH of the rain water falls below 5.6, it is called acid rain. When it flows into the rivers, it lowers the pH of river water; hence the survival of the aquatic life becomes difficult. When such water is absorbed by the soil, the soil becomes acidic and unfertile. The acidity is neutralized by adding slaked lime (Calcium Hydroxide). (iii) Honey bee or ant injects methanoic acid (formic acid) into our body causing pain and irritation. Use of mild base like baking soda on the stung area gives relief as it neutralizes the acid. (iv) Leaves of nettle plants contain methanoic acid causing pain and irritation. Rubbing with the leaf of dock plant (which also grows nearby) gives relief as it contains mild base and neutralizes the acid. |
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783. |
Which of the following cations is not hydrolyzed in aqueous solution ? (i) `Ba^(2+)` , (ii) `Ca^(2+)` (iii) `Na^(+)` , (iv) `K^(+)`A. (i), (ii)B. (iii), (iv)C. (i), (ii), (iii), (iv)D. (i), (ii), (iii) |
Answer» Correct Answer - C Salts formed by the reactions between acids and bases (in denfinite proporties) undergo dissociation in water. The cation/anions formed on dissociation of salts either exist as hydrated ions in aqueous solutions or interact with water to reform the corresponding acids/bases depending upon the nature of salts. The latter process of interaction between water and cations/anions or both of salts is called hydrolysis. The pH of the solution gets a affected by this interaction. The cations of strong bases such as `Na^(+), K^(+), Ca^(2+)`, and `Ba^(2+)` simply get hydrated but do not hydrolyze. |
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784. |
Which of the anions is not hydrolyzed in aqueous solution ? `CI^(-)` , (ii) `NO_(3)^(-)` (iii) `Br^(-)` , (iv) `CIO_(4)^(-)`A. (i), (ii), (iii), (iv)B. (ii), (iii), (iv)C. (i), (ii), (iii)D. (ii), (iv) |
Answer» Correct Answer - A The anions of strong acids `(CI^(-), Br^(-), NO_(3)^(-)`, and `SO_(4)^(2-))` simply get hydrated but do not hydrolyze. |
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785. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures of acids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is `pH=pK_(a) + log. (["Salt"])/(["Acid"])` a similar equation is used for basic buffers. The pH of aqueous solution of single salts is calculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is `pH = 7-(1)/(2) pK_(b)-(1)/(2) log c` For weak acids and bases used by a chemist, data are given below: `K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5)` Also logarithmic values of some numbers are given below : log 1.8 = 0.2553, log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH value in each of the following cases. 100 mL of 0.10 M `NH_(4)OH` mixed with 100 of 0.05 M HCl solutionA. 6.25B. 7.25C. 8.25D. 9.25 |
Answer» Correct Answer - D 100 ml of 0.10 M `NH_(4)OH` = 10 mmol of `NH_(4)OH` 100 mL of 0.05 M HCl = 5 mmol of HCl 5 mmol of HCl will react with 5 mmol of `NH_(4)OH` to form 5 mmol of `NH_(4)Cl` `NH_(4)OH ` left = 5 mmol , `NH_(4)Cl` formed = 5 mmol Thus, it is a basic buffer with `[NH_(4)OH]=[NH_(4)Cl]` `pOH = pK_(b) + log .(["Salt"])/(["Base"])` `=-log (1.8xx10^(-5))=4.75` `:. pH = 14 - 4.75 9.25` |
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786. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures of acids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is `pH=pK_(a) + log. (["Salt"])/(["Acid"])` a similar equation is used for basic buffers. The pH of aqueous solution of single salts is calculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is `pH = 7-(1)/(2) pK_(b)-(1)/(2) log c` For weak acids and bases used by a chemist, data are given below: `K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5)` Also logarithmic values of some numbers are given below : log 1.8 = 0.2553, log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH value in each of the following cases. 100 mL of 0.10 M NaOH mixed with 100 ml of 0.05 M `CH_(3)C O O H` solutionA. 10.4B. 11.4C. 12.4D. 13.4 |
Answer» Correct Answer - C 100 ml of 0.1 M NaOH = 10 mmol of NaOH 100 mL of 0.05 M `CH_(3)CO OH` = 5 mmol of `CH_(3)CO OH` 5 mmol will react with 5 mmol of NaOH to form 5 mmol of `CH_(3)CO Ona` NaOH left = 5 mmol. Volume of the solution = 200 ml `:. [NaOH]=(5)/(200) M = 0.025M` or `[OH^(-)]=0.025M` or `[OH^(-)]=0.025M` `:. [H^(+)]=10^(-14)//0.025=4.0xx10^(-13)`, pH = 13- 0.602 = 12.398 `~=` 12.4 |
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787. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures of acids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is `pH=pK_(a) + log. (["Salt"])/(["Acid"])` a similar equation is used for basic buffers. The pH of aqueous solution of single salts is calculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is `pH = 7-(1)/(2) pK_(b)-(1)/(2) log c` For weak acids and bases used by a chemist, data are given below: `K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5)` Also logarithmic values of some numbers are given below : log 1.8 = 0.2553, log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH value in each of the following cases. 100 mL of 0.10 M NaOH mixed with 100 mL 0.10 M `CH_(3)CO OH` solutionA. 5.72B. 6.72C. 7.72D. 8.72 |
Answer» Correct Answer - D 100 ml of 0.1 M NaOH = 10 mmol of NaOH 100 ml of 0.10 M `CH_(3)CO OH` = 10 mmol of `CH_(3)CO OH` They will react completely to form 10 mmol of `CH_(3)CO ON a` Volume of solution = 200 mL `:. [CH_(3)CO ON a]=(10)/(200)= 0.05 M` As `CH_(3)CO ON a` is a salt of weak acid with strong base, `pH= 7 +(1)/(2) pK_(a)+(1)/(2) log c` `= 7 + (1)/(2) (4.74)+(1)/(2) log 0.05` `=7+2.37 + (1)/(2) (-1.301)=8.72` |
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788. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures of acids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is `pH=pK_(a) + log. (["Salt"])/(["Acid"])` a similar equation is used for basic buffers. The pH of aqueous solution of single salts is calculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is `pH = 7-(1)/(2) pK_(b)-(1)/(2) log c` For weak acids and bases used by a chemist, data are given below: `K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5)` Also logarithmic values of some numbers are given below : log 1.8 = 0.2553, log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH value in each of the following cases. 100 mL of 0.05 M NaOH mixed with 100 ml of 0.10 M `CH_(3)CO OH` solutionA. 3.75B. 4.75C. 5.75D. 6.75 |
Answer» Correct Answer - B 100 ml of 0.05 M NaOH = 5 mmol of NaOH 100 mL of 0. 1 M `CH_(3)CO OH= 10 ` mmol of `CH_(3)CO OH` 5 mmol of NaOH will react with 5 mmol of `CH_(3)CO OH` to form 5 mmol of `CH_(3)CO ON a` `CH_(3)CO OH ` left = 5 mmol, `CH_(3)CO ON a` formed = 5 mmol Total volume = 200 mL `:. [CH_(3)CO OH]=[CH_(3)CO ON a]` `=(5)/(200) M = 0.025 M` As it is a buffer of weak acid and its salt with strong base, `pH = pK_(a)+log. (["Salt"])/(["Base"])` `=-log (1.8xx10^(-5))+log 1` `=5- (0.255)=4.745` |
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789. |
The dissociation constant of monobasic acids A.B and C are `10^(-4),10^(-6)and 10?^(-10)` respectively, The concentration of each monobasic acid is 0.1 M Which of the following has been arranged in inc reasing order of pH ?A. `CltBltA`B. `AltBltC`C. `Blt CltA`D. `BltAltC` |
Answer» Correct Answer - B |
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790. |
When `PCl_(5)` is heated in a closed vessel at 575 K, the total pressure at equilibrium is found to be 1 atm and partial pressure of `Cl_(2)` is found to the 0.324 atm . Calculate the equilibrium constant `(K_(p))` for the decomposition reaction. |
Answer» Correct Answer - `0.298` ` PCl_(5) hArr PCl_(3) + Cl_(2)` `" At eqm., " p_(Cl_(2)) = p_(pCl_(3))= 0*324 " atm . Hence " p_(PCl_(5)) = 1- (0*324 + 0*324) = 0*352 " atm"` `K_(p) = (p_(PCl_(3)) xxp_(Cl_(2)))/(p_(PCl_(5)))=(0*324 xx 0*324)/(0*352)= 0*298 " atm "` |
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791. |
Ammonium carbamate when heated to `200^(@) C ` gives a mixture of `NH_(3) and CO_(2)` vapours with a density of `16*0` . What is the degree of disociation of ammonium carbamate ?A. `3//2`B. `1//2`C. 2D. 1 |
Answer» Correct Answer - D `{: (,NH_(2)COONH_(4),to,2NH_(3),+,CO_(2)), ("Intial" ,1"mole",,,,), ("After disso.",1-alpha,,2alpha,,alpha ),(,,,,,"Total"=1+2 alpha):}` Theoretical density `(D) propto 1/V ` Observed density ` (d) propto 1/((1+2alpha)V)` ` :. D/d = 1+ 2 alpha ` or ` alpha = 1/2 ((D-d)/d) = 1/2 ((48-16*0)/16*0) = 1*0 ` |
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792. |
In an experiment , 2 moles of HI are taken into an evacuated 10.0 litre container at 720 K. The equilibrium constant equals to 0.0156for the gaseous reaction, `2 HI (g) hArr H_(2) (g) + I_(2)(g).` find equilibrium concentration of ` HI (g) , H_(2) (g) , I_(2)(g).` |
Answer» Correct Answer - `HI = 1.6 " mole " H_(2)= 0.2 " mole " I_(2) = 0.2 " mole "` Suppose x moles of `H_(2)` are formed at eqm. Then concentrations at equilibrium will be : `[HI] = (2-2 x ) //10,[H_(2)] = x // 10 and [I_(2)] = x//10 " mol " L^(-1).` |
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793. |
A mixture of `NO_(2) and N_(2)O_(4)` has a vapour density of `38*3` at 300 K . What is the number of moles of `NO_(2)` in 100 g of the mixture ?A. `0*043`B. `4*4`C. `3*4`D. ` 0*437` |
Answer» Correct Answer - D Suppose `NO_(2) = x g . " Then " N_(2) O_(4) = ( 100 - x) g ` `" Moles of " NO_(2)= x/46,"Moles of " N_(2)O_(4)=(100-x)/92` `"Mole fraction of "NO_(2)(x//46)/(x//46 + ( 100 - x ) //92)` `= x/46 xx92/(100+x)=(2x)/(100+x)` Mole fraction of `N_(2)O_(4) = 1 - (2x)/(100+x) = (100 - x)/(100+x) ` Molar mass of mixture `= (2x)/(100 + x) xx 46 + ( 100-x)/(100+x) xx92 = 9200/(100+x) ` ` :. 9200/(100+x) = 2 xx 38*3 = 76*6` or` 76*6 x = 9200 = 1540 or x = 20*10 g ` ` :." Moles of " NO_(2) = (20*10)/46 = 0. 437` |
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794. |
In the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as –(a) If both assertion and reason are true, and reason is the true explanation of the assertion.(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.(c) If assertion is true, but reason is false.(d) If both assertion and reason are false Assertion (A). Kp can be equal to or less than or even greater than the value of Kc.Reason (R). Kp = Kc(RT)Δn.Relation between Kp and Kc depends on the change in the number of moles of gaseous reactants and products. |
Answer» (a) If both assertion and reason are true, and reason is the true explanation of the assertion. |
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795. |
will the pH of water be same at `4^(@)C` and `25^(@)C` ? Explain. |
Answer» No, pH of water is not same at `4^(@)C and 25^(@)C`. This is because with increase in temperature, dissociation of `H_(2)O` molecules increases. Hence, `[H^(+)]` will increase, i.e., pH decreases. Thus, pH at `4^(@)C` will be more than at `25^(@)`C. | |
796. |
Which one of the following is the correct statement ?A. `HC_(3)^(-)` is the conjugate base of `CO_(3)^(2-)`B. `NH_(2)^(-)` is the conjugate acid of `NH_(3)`C. `NH_(3)` is the conjugate base of `NH_(2)^(-)`D. `H_(2)CO_(3)` is the conjugate base of `HCO_(3)^(-)` |
Answer» Correct Answer - C `{:(HCO_(3)^(-),underset(.^(-)H^(+))(to),CO_(3)^(2-)",",NH_(3),underset(.^(-)H^(+))(to),NH_(2)^(-)),(underset("acid")("Conjugate"),,underset("base")("Conjugate"),underset("acid")("Conjugate"),,underset("base")("Conjugate")),(H_(2)SO_(4),underset(.^(-)H^(+))(to),HSO_(4)^(-)",",H_(2)CO_(3),underset(.^(-)H^(+))(to),HCO_(3)^(-)),(underset("acid")("Conjugate"),,underset("base")("Conjugate"),underset("acid")("Conjugate"),,underset("base")("Conjugate")):}` |
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797. |
Conjugate base of `H_(2)` isA. `H_(3)^(+)`B. `H_(3)^(-)`C. `H^(+)`D. `H^(-)` |
Answer» Correct Answer - D Conjugate base = Conjugate acid `-H^(+)=H_(2)-H^(+)=H^(-)` |
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798. |
For an aqueous solution of `NH_(4)Cl`, prove that `[H_(3)O^(+)]=sqrt(K_(h)C)`. |
Answer» For salt of strong acid and weak base, `[H_(3)O^(+)]=sqrt((K_(w)C)/(K_(b)))` Substituting `(K_(w))/(K_(b))=K_(h)`, we get `[H_(3)O^(+)]=sqrt(K_(h)C)`. |
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799. |
The pH of an enzyme catalysed reaction has to be maintained between 7 and 8. What indicator should be used to monitor and a control the pH ? |
Answer» Bromothymol blue or Phenol red or Cresol red . | |
800. |
What common salt is added to precipitate out soap from the solution during its manufacture? |
Answer» Soap is sodium salt of higher fatty acids (RCOONa). On adding common salt, Na+ ion concentration increased. Hence, the equilibrium RCOONa (s) ----- RCOO- + Na + shifts in the backward dirction, i,e, soap precipitates out. |
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