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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.
| 1101. |
Aquenous solution of which of the following compounds is the best conductor of electric current ?A. Ammonia, `NH_(3)`B. Fructose, `C_(6)H_(12)O_(6)`C. Acetic acid, `C_(2)H_(4)O_(2)`D. Hydrochloric acid HCl |
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Answer» Correct Answer - D Fructose is a non-electrolyte, `NH_(4)OH` and `CH_(3)COOH` are weak electrolytes, HCI is a strong electrolyte. Hence, HCI dissociates completely in aqueous solution and, therefore is the best conductor of electricity . |
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| 1102. |
Assertion : For the reaction : `A(g) +B(g) hArr C (g) +D (g)` at a given temperature , there will be no effect of the addition inert gas at constant pressure or at constant volume. Reason : For the reaction `Delta^(n) =0.` Therefore there is no effect of addition of inert gas either at constant volume or at constant pressure.A. if both assertion and reason are correct and reason is correct explanation for assertion.B. if both assertion and reason are correct but reason is not correct explanation for assertion.C. if assertion is correct but reason is incorrect.D. if assertion and reason boht are incorrect. |
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Answer» Correct Answer - A Reason is the correct explanation for assertion. |
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| 1103. |
Consider the following equilibrium system: `2SO_2(g)+O_2(g) hArr 2SO_3(g)` Some inert gas is added to the above system at constant volume. Predict which of the following is true?A. More of `SO_3` is produced.B. Less `SO_2` is produced.C. Addition of inert gas does not affect equilibrium.D. System moves to new equilibrium position which can not be predicted theoretically. |
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Answer» Correct Answer - C If the gas that is introduced (at constant volume) is notinvolved in the reaction, the partial pressure of each reacting has remains constant. So the system remains at equilibrium. However, the total pressure of a gaseous system is raised by pumping an inert gas. |
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| 1104. |
Which of the following are Lewis acids?H2O, BF3, H+, NH4+ |
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Answer» BF3, H+ and NH4+ are Lewis acids as these can accept electron pairs due to deficiency of electrons. |
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| 1105. |
Hydrochloric acid present in the gastric juice is secreted by the lining of our stomach in a significant amount of __________ L `day^(-1)` .A. `1.2-1.5`B. `0.5-1.0`C. `1.5-2.0`D. `1.0-2.0` |
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Answer» Correct Answer - A It is essential for difestive process. |
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| 1106. |
What will be the effect on equilibrium if (i) a catalyst is added to it (ii) An inert gas is added at constant volume |
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Answer» (i) The state of equilibrium is not disturbed on adding catalyst rather it is attained quickly. (ii) No effect on state of equilibrium |
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| 1107. |
What is the effect of increasing pressure on the equilibrium state of the reaction?N2 (g) + 3H2 (g) → 2NH3 (g) + Heat |
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Answer» For the reaction, N2 (g) + 3H2 (g) → 2NH3 (g) + Heat On increasing the pressure, the reaction will proceed in forward direction. This results in increase in yield of ammonia. |
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| 1108. |
The solubility product of AgCl is `4.0xx10^(-10)` at 298 K. The solubility of AgCl in 0.04 m `CaCl_(2)` will beA. `2.0xx10^(-5) m`B. `1.0xx10^(-4)m`C. `5.0xx10^(-9)m`D. `2.2xx10^(-4)`m |
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Answer» Correct Answer - C If x is the solubility of AgCl in 0.04 m `CaCl_(2)`, then `[Ag^(+)]= x "mol" L^(-1), [Cl^(-1)]=(0.04xx2)+x ~~0.8 m` `:. x (0.08 ) 4xx10^(-10) "or" x = 5.0 xx 10^(-9) m` |
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| 1109. |
What is the effect of increasing pressure on the equilibrium? N2 + 3H2 ⇌ 2NH3? |
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Answer» Equilibrium will shift in the forward direction forming more of ammonia. |
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| 1110. |
What is the effect on the value of equilibrium constant on adding catalyst? |
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Answer» No effect on the value of equilibrium constant on adding catalyst. |
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| 1111. |
A reaction `A (g) + B (g) hArr 2 C (g) ` is in equilibrium at a certain temperature. Can we increase the amount of products by (i) adding catalyst (ii) increasing pressure ? |
| Answer» (i) No, because catalyst does not disturb the state of equilibrium (ii) No, because `n_(p) = n_(r)` | |
| 1112. |
Which of the following is not a Lewis acid ?A. `AICI_(3).6H_(2)O`B. `AICI_(3)`C. `SnCI_(4)`D. `FeCI_(3)` |
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Answer» Correct Answer - A A Lewis acid is characterized by its ability to accept at least one lone pair using an empty orbital of right energy. `AlCl_(3), SnCl_(4)`, and `FeCl_(3)` act as Lewis acids because of the ability of their central atoms to accept lone pairs in their empty orbitals. `AlCl_(3).6H_(2)O` or `[Al(H_(2)O)_(6)]Cl_(3)` cannot act as a Lewis acid because the central ion `Al^(3+)` is no longer in a position to accept the lone pair as its coordination number is fulfilled. |
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| 1113. |
Which of the following pairs consitutes buffer?A. `HNO_(3)` and `NH_(4)NO_(3)`B. HCI and KCIC. `HNO_(2)` and `NaNO_(2)`D. `NaOH` and `NaCI` |
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Answer» Correct Answer - C It is an acidic buffer. `HNO_(2)` is a weak acid and `NaNO_(2)` is the salt of `HNO_(2)` with a strong base NaOH. |
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| 1114. |
Calculate the pH value of `(i) 10^(-2) "molar" HNO_(3) " solution " (ii) " 0.03 N HCl solution " (iii) 0.001 N H_(2)SO_(4) "solution "`. |
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Answer» (i) `HNO_(3)` completely ionizes as : `HNO_(3)+H_(2)O rarr H_(3)O^(+)+NO_(3)^(-)` `:. [H_(3)O^(+)]=10^(-2)M` (Given) `pH=-log[H_(3)O^(+)]=-log(10^(-2))=-(-2 log 10) = 2` (ii) HCl completely ionizes as : `HCl + H_(2)O rarr H_(3)O^(+) + Cl^(-)` `:. [H_(3)O^(+)]=[HCl]=0.03 N ` (Given) `=3xx10^(-2)N = 3xx10^(-2)M ` (HCl is monobasic, Eq. mass = Mol. mass) `:. pH = - log [H_(2)O^(+)]=- [log 3 xx10^(-2)]=-(log 3 + log 10^(-2))=- (0.4771-2) = 1.5229` (ii) `H_(2)SO_(4) ` completely ionizes as : `H_(2)SO_(4) + 2H_(2)O rarr 2H_(3)O^(+)+SO_(4)^(2-)` `{:(,[H_(3)O^(+)]=2xx[H_(2)SO_(4)],,[1 "molecule of" H_(2)SO_(4) "gives"2H_(3)O^(+) "ions"]^(**),,),("But",H_(2)SO_(4)=0.001 N = 0.001 xx49 "g/litre",,("Eq. mass of " H_(2)SO_(4)=49),,),(," "=(0.001xx49)/(98) "moles/litre",,("Mol. mass of " H_(2)SO_(4)=98),,),(," "=0.0005 M ,,("Directly, Molarity"("Normality")/("Basicity")),,):}` `:. [H_(3)O^(+)]=2xx[H_(2)SO_(4)]=2xx0.0005M = 0.001 M = 10^(-3)M` `:. pH = - log [H_(3)O^(+)]=-log (10^(-3))=3` |
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| 1115. |
The pH of 0.001 M `Ba(OH)_(2)` solution will beA. 2B. 8.4C. 11.3D. 2.7 |
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Answer» Correct Answer - C `Ba(OH)_(2)hArrBa^(2+)+2OH^(-)` `[OH^(-)]=2xx1xx10^(-3)M` `pOH=-log[OH^(-)]=-log(2xx10^(-3))=2.7` `pOH+pH=14pH=14-2.7=11.3` |
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| 1116. |
The equilibrium constant `K_p` for the reaction `AhArr2B` is related to the degree of dissociation `(alpha)` of A and total pressure `P` asA. `(4alpha^2P)/(1-alpha)`B. `(4alpha^2P^2)/(1-alpha)`C. `(4alpha^2P)/(1-alpha^2)`D. `(4alpha^2P^2)/(1-alpha^2)` |
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Answer» Correct Answer - C `{:(,A hArr 2B),("Initial moles","1 0"),("Change",-alpha+2alpha),("Equilibrium moles",bar(1-alpha " 2"alpha)):}` Partial pressure =(Mole fraction)(Total pressure) Mole fraction = Moles/Total moles `n_A=1-alpha`, `n_B=2alpha`, and `n_t=(1-alpha)+(2alpha)=1+alpha` `:. P_A=((1-alpha)/(1+alpha))P`, `P_B=((2alpha)/(1+alpha))P` |
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| 1117. |
Caculate the pH of 0.001 N `H_(2)SO_(4)` solution. |
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Answer» Correct Answer - 3 `H_(2)SO_(4)` is a strong acid and it is completely ionised in the aqueous solution as : `H_(2)SO_(4)overset((aq))(to) H_(3)O^(+) +SO_(4)^(2-) (aq)` As `H_(2)SO_(4)` is a dibasic acid , the molarity of the solution is half of its normality or 0.0005 M `[H_(3)O^(+)] =2xx 0.0005 =0.001 M =10^(-3) M` `pH =- log [H_(3)O^(+)]=- log [(10^(-3))] =(-3) (-log 10) =3` |
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| 1118. |
The relation between `K_p` and `K_x` isA. `K_p=K_x((P)/(sumn))`B. `K_p=K_x(P)^(-Deltan)`C. `K_p=K_x(P)^(Deltan)`D. `K_p=K_x(RT)^(Deltan)` |
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Answer» Correct Answer - C Consider the equilibrium equation `aA+bB hArr cC+dD` `K_P=(P_C^cP_D^d)/(P_A^aP_B^b)` For any gas, partial pressure P of gas `=("Mole fraction " chiof " gas")xx("Total pressure " P_t)` `:. K_p=(chi_C^c chi_D^dP_t^(c+d))/(chi_A^achi_B^dP_t^(a+b))=K_chiP_t^((c+d)-(a+b))` `=K_(chi)(P_t)^(Deltan)` |
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| 1119. |
An equilibrium mixture for the reaction `2H_(2)S(g) hArr 2H_(2)(g) + S_(2)(g)` had 1 mole of `H_(2)S, 0.2` mole of `H_(2)` and 0.8 mole of `S_(2)` in a 2 litre flask. The value of `K_(c)` in mol `L^(-1)` isA. 0.004B. 0.08C. 0.016D. 0.16 |
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Answer» Correct Answer - C `K_(c) =[[H_(2)]^(2)[S_(2)]]/[[H_(2)S]^(2)]` `[H_(2)] =(2xx0.2)/(2) =0.2 " mole"//L` `[S_(2)]=(0.8)/(2) =0.4 "mole"//L ,[H_(2)S]=1 "mole"//L` `:. K_(c) =[[0.2]^(2)[0.4]]/(1) =0.016 "mole"//L` |
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| 1120. |
The decreasing order of strength of the bases, `OH^(-), NH_(2)^(-), H-C-=C^(-)` and `CH_(3)-CH_(2)^(-)`:A. `CH_(3) -CH_(2)^(-) gt NH_(2)^(-) gt H-C -= C^(-) gt OH^(-)`B. `H-C -= C^(-) gt CH_(3) -CH_(2)^(-) gt NH_(2)^(-) gt OH^(-)`C. `OH^(-) gt NH_(2)^(-) gt H-C -= C^(-) gt CH_(3) -CH_(2)^(-)`D. `NH_(2)^(-) gt H -C -= C^(-) gt OH^(-) gt CH_(3) -CH_(2)^(-)` |
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Answer» Correct Answer - B the conjugate acids of the bases `OH^(-),NH_(2)^(-), H-C-=C^(-) " and "CH_(3)-CH_(2)^(-)` are respectively `H-OH, NH_(3)^(2), H-C-=CH " and " CH_(3)CH_(3).` Their acidic nature is in the order : `HOH gt HC-=CH gt NH_(3) gt CH_(3)-CH_(3)` Now a strong acid has weak conjugate base. Thus the strength of the bases is in the order `CH_(3)-CH_(2)^(-) gt HC -=C^(-) gt OH^(-)` |
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| 1121. |
If concentration of OH- ions in the reaction Fe(OH)3(s) ⇋ Fe3+(aq) + 3OH- (aq) is decreased by 1/4 times, then equilibrium concentration of Fe3+ will increase by –(a) 8 times (b) 16 times (c) 64 times (d) 4 times |
| Answer» (c) 64 times | |
| 1122. |
In which of the following equilibrium, Kc and Kp are not equal?(a) 2C(s) + O2(g) ⇋ 2CO2(g)(b) 2NO(g) ⇋ N2(g) + O2(g)(c) SO2(g) + NO2(g) ⇋ SO3(g) + NO(g)(d) H2(g) + I2(g) ⇋ 2HI(g) |
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Answer» (a) 2C(s) + O2(g) ⇋ 2CO2(g) |
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| 1123. |
The hydronium ion concentration of a solution is `1.3 xx 10^(-5)M`. Find out its pH value. Predict the nature of the solution. |
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Answer» Correct Answer - `4.88` Acidic `[H_(3)O^(+)]=1.3 xx 10^(-5)M` `pH =- log [H_(3)O^(+)] =- log (1.3 xx 10^(-5))` `=(5 log 10 -log 1.3) =(5-0.114) =4.88` Since the pH of the solution is less than 7 , it is acidic in nature . |
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| 1124. |
Calculate the `pOH` of solution at `25^(@)C` that contains `1xx10^(-10) M` of hydronium ions, i.e., `H_(3)O^(+)`A. `1.000`B. `7.000`C. `4.000`D. `9.000` |
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Answer» Correct Answer - C `C_(H_(3)O^(+))=10^(-pH) mol L^(-1)` `=10^(-10)mol L^(-1)` `:. pH = 10` Now, for any aqueous solution at `25^(@)C`, `pH+pOH=14` `:. pOH=14-pH` `=14-10` `=4` |
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| 1125. |
The dissociation constants for acetic acid and HCN at `25^(@)C` are `1.5xx10^(-5)` and `4.5xx10^(-10)` , respectively. The equilibrium constant for the equilibirum `CN^(-) + CH_(3)COOHhArr HCN + CH_(3)COO^(-)` would beA. `3.0xx10^(-5)`B. `3.0xx10^(-4)`C. `3.0xx10^(4)`D. `3.0xx10^(5)` |
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Answer» Correct Answer - C We are given `CH_(3)COOH hArr CH_(3)COO^(-)+H^(+), K_(1)=1.5xx10^(-5)` `HCN hArr H^(+)+CN^(-), K_(2)= 4.5xx10^(-10)` If we reverse the second equilibrium expression and add it to the first one, we get `CN^(-)+CH_(3)COOH hArr CH_(3)COO^(-)+HCN` `:. K_(eq)=K_(1)xx(1)/(K_(2))=(1.5xx10^(-5))/(4.5xx10^(-10))` `=0.33xx10^(5)` `=3.3xx10^(4)` Note that equilibrium constant expressions are multiplied when we add the expressions. When w reverse the expression, the new equilibrium constant is always the reciprocal of the previous one. |
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| 1126. |
The dissociation constant of acetic acid and HCN at 250C are 1.5 × 10-5 and 4.5 × 10-10 respectively. The equilibrium constant for the equilibriumCN- + CH3COOH ⇋ HCN + CH3COO- would be –(a) 3.0 × 10-5 (b) 3.0 × 10-4 (c) 3.0 × 104 (d) 3.0 × 105 |
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Answer» (c) 3.0 × 104 |
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| 1127. |
The concentration of hydronium ions in a cup of black coffee is `1.3xx10^(-5)` M. Find the pH of the coffee. Is this coffee acidic or alkaline ? |
| Answer» Correct Answer - 4.89, acidic | |
| 1128. |
Calculate the pH of a solution obtained by mixing of 100 ml of 0.1 M HCI and 100 ml of 0.2 M NH3 ? Kb, for NH3 = 1.8 x 10-5 ? |
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Answer» 100 ml of 0∙1 M HCl + 100 ml of 0∙2 M NH3 Kb for NH3 = 1.8 x 10-5 100 × 0∙1 = 10 m mol of HCl 100 × 0∙2 = 20 m mol of NH3 NH3 (1 mol) + HCl (1 mol) → NH4Cl (1 mol) i.e., 10 m mol 10 m mol 10 m mol NH3 left unreacted = 10 m mol Volume of solution = 200 ml Molarity of NH3 = \(\frac{10}{200}\) M = 0 ∙ 05 M [NH4+] = \(\frac{10}{200}\) M = 0 ∙ 05 M pOH = pKb + log\(\frac{[Salt]}{[Acid]}\) = -log (1.8 x 10-5) + log \(\frac{0.05}{0.05}\) = 4∙74 pH = 14 – pOH = 14 − 4∙74 = 9∙26. |
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| 1129. |
pH of black coffee is 5.0 at 25°C. Is black coffee acidic or basic? |
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Answer» Since, the pH of black coffee is less than 7 i.e. 5.0 so, it is acidic. |
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| 1130. |
What will be the pH of 0.01 M HCI solution at 25°C? |
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Answer» Given, [H+] = 0.01 M = 1 × 10-2 M \(\because\) pH = -log10 [H+] = -log10 [10-2 ] = 2 log 10 = 2 × 1 = 2 |
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| 1131. |
`10^(-5)M HCI` solution at `25^(@)C` is dilluted 1000 times. The pH of the diluted solution willA. be equal to 8B. remain unchangedC. lie between 5 and 6D. lie between 6 and 7 |
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Answer» Correct Answer - D According to dilution equation, `M_(i)V_(i)=M_(f)V_(f)` `(10^(-5))(x)=M_(f)(1000 x)` `M_(f)=10^(-8)` Therefore, when `10^(-5)` M HCI is diluted 1000 times, the concentration of the resulting soluting will be `10^(-8)`M. Since HCI is a strong acid, the pH of the diluted solution is expected to be 8. This result cannot be accepeted because this value comes in the basic range. pH can be brought down in the acidic range if more `H^(+)` ions are considered in calculation. The other source of `H^(+)` ions is `H_(2)O.H^(+)` ion concentrations form the acid and water are comparable. Moreover, the concentration of `H^(+)` ions form `H_(2)O` is no longer `10^(-7)` M as the equilibrium of water is disturbed on the addition of HCI. Let us assume that when equilibrium is restableished, the concentration of `H^(+)` ions form `H_(2)O` is x mol `L^(-1)` `2H_(2)O(aq.) hArr H^(+)underset(x)((aq.))+OH^(-)underset(x)((aq.))` Total `H^(+)` concentration will be `C_(H^(+))` form `H_(2)O` and `C_(H^(+))` form HCI. `K_(w)=C_(H^(+))C_(OH^(-))` `1xx10^(-14)=(x+10^(-8))(x)` `x^(2)+10^(-8)x-10^(-14)=0` Solving the quardatic equation gives `x=9.5xx10^(-8)` Thus, total `C_(H^(+))=9.5xx10^(-8)+10^(-8)` `= 1.05xx10^(-7) mol L^(-1)` `pH= -(1ogC_(H^(+)))/(mol L^(-1))` `= -1og (1.05xx10^(-7))` `=6.98` Now, remember the following guidellince while calculating the pH of an acid: (i) `C_(H^(+)=C_(H^(+))` form acid only if `C_("acid")` is greater than `10^(-5) M`. (ii) `C_(H^(+))=C_(H^(+))` form acid `+ C_(H^(+))` form water if `C_("acid")` lies between `10^(-8)` M and `10^(-5)` M. (iii) `C_(H^(+))=C_(H^(+))` from water only if `C_("acid")` is far less than `10^(-8)` M. |
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| 1132. |
Calculate pH of 0.01 M HCI solution. |
| Answer» `pH =- log [H^(+)] =- log [10^(-2)] =2` | |
| 1133. |
What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?A. `K_(c)=[O_(2)]^(5)`B. `K_(c)=[P_(4)O_(10)]//[P_(4)][O_(2)]^(5)`C. `K_(c)=[P_(4)O_(10)]//4[P_(4)][O_(2)]`D. `K_(c)=(1)/([O_(2)]]^(5)` |
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Answer» Correct Answer - D is the correct answer. |
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| 1134. |
What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?A. `K_(eq)=[O_(2)]^(5)`B. `K_(eq)=(1)/(5)([P_(4)O_(10)])/([P_(4)][O_(2)])`C. `K_(eq)=(1)/([O_(2)]^(5))`D. `K_(eq)=([P_(4)O_(10)])/([P_(4)][O_(2)]^(5))` |
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Answer» Correct Answer - C For pure solids such as `P_(4)O_(10)` and `P_(4)`, the active mass (expressed through density) is constant at a given temperature. For mathematical convenience, it is taken as unity (1). Thus, `K_(eq)= (1)/([O_(2)]^(5))` |
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| 1135. |
The dissociation of phosgene gas `(COCl_(2))` is represented as : `COCl_(2) (g) hArr CO(g) + Cl_(2)(g)` When the mixture of these three gases is compressed at constanat temperature , what happens to (i) the amount of CO in the mixture (ii) the partial pressure of `COCl_(2)` (iii) the equilibrium constant for the reaction ? |
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Answer» The dissociation reaction proceeds with the increase in the number of moles per unit volume . When the mixture of the gases is compressed at constant temperature, this means that the number of moles per unit volume further increase . In order to undo the effect, the equilibrium must shift to the left As a result, `COCl_(2)(g)hArr CO(g) +Cl_(2)(g)` (i) the amount of `CO(g)` in the mixture decreases. (ii) The partial pressure of `COCl_(2)(g)` increase. (iii) Since the partial pressure of both CO(g) and `Cl_(2)(g)` decrease and that of `COCl_(2)(g)` increases, therefore the value of `K_(p)` decreases. |
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| 1136. |
What is the `[OH^(-)]` in the final solution prepared by mixing `20.0 mL` of `0.050 M HCl` with `30.0 mL` of `0.10 M Ba(OH)_(2)`?A. 0.40 MB. 0.0050 MC. 0.12 MD. 0.10 M |
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Answer» Correct Answer - D Millimoles of `H^(+)` produced `=20 xx 0.05 =1` Millimoles of `OH^(-)` produced `=30 xx 0.1 xx 2` , (Each `Ba(OH)_(2)` given two `OH^(-)` ions) =6 Millimoles of `OH^(-)` ions in the remaining solution =6 -1 =5 Total volume of solution =20 + 30 =50 mL `[OH^(-)] = (5)/(50) =0.1 M` |
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| 1137. |
What is the `[OH^(-)]` in the final solution prepared by mixing `20.0 mL` of `0.050 M HCl` with `30.0 mL` of `0.10 M Ba(OH)_(2)`?A. `0.40M`B. `0.050M`C. `0.12M`D. `0.10M` |
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Answer» Correct Answer - D Normality = n-factor x Molarity. For HCl, n-factor is 1 and for `Ba(OH)_(2)`, n-factor is 2. Thus, `0.05 0M HCl = 0.50N HCl` `0.10 M HCI = 0.20 N Ba(OH)_(2)` Milliequivalents `(H^(+))` = Milliequivalents (HCI) `=(20 ml)(0.050 N)` `=1.0` Millequivalents `(OH^(-))` = Millequivalents `[(Ba(OH)_(2)]` `=(30 ml)(0.20 N)` `=6.0` 1 meq of `H^(+)` will neutralized 1 meq of `OH^(-)` . Thus, 5 meqs of `OH^(-)` are left. Since the total volume is `20+30=50 mL` , we have `N(OH^(-))=(meqs)/(mL)=(5)/(50)"meq mL"^(-1)` `=0.1` meq `mL^(-1)` Since `OH^(-)` is univalent, its normality and molarity are the same. Thus, `M(OH^(-))=0.1 mol L^(-1)` Alternatively, `Ba(OH)_(2)+2HCl rarr BaCl_(2)+2H_(2)O` According to equation, 2 millimoles of HCl neutralize 1 millimole of `Ba(OH)_(2)`. Thus, 1 mmol of HCI will neutralized `0.5` mmol of `Ba(OH)_(2)`. As a result, `3-0.5=2.5` mmol of `Ba(OH)_(2)` are left unneutralized. `M[Ba(OH)_(2)]= (mmol)/(mL)=(2.5)/(50)=0.05` `M(OH^(-))=2 M [Ba(OH)_(2)]` `= 2(0.05)=0.1M` Note that 1 unit of `Ba(OH)_(2)` yields two units of `OH^(-)` ions. |
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| 1138. |
Give reason for the following:Zinc is not precipitated as Zn (OH)2 on adding NH4 OH to a zinc salt solution containing NH4Cl. |
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Answer» NH4CL suppresses the ionization of NH4OH due to common ion effect. Hence, the concentration of OH- decreases such that the ionic product [Zn2+] [OH-]2 does not exceed solubility product. Hence, Zn()H)2 is not precipitated. |
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| 1139. |
A mixture of `1.57 mol` of `N_(2), 1.92 mol` of `H_(2)` and `8.13 mol` of `NH_(3)` is introduced into a `20 L` reaction vessel at `500 K`. At this temperature, the equilibrium constant `K_(c )` for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `1.7xx10^(2)`. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?A. ForwardB. BackwardC. At equilibriumD. Data is insufficient |
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Answer» Correct Answer - B The reaction is : `N_(2(g))+3H_(2(g))hArr2NH_(3(g))` `Q_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))` `=(((8.13)/(20)"mol L"^(-1))^(2))/(((1.57)/(20)"mol L"^(-1))((1.92)/(20)"mol L"^(-1))^(3))=2.38xx10^(3)` As `Q_(c)!=K_(c)`, the reaction mixture is not in equilibrium. As `Q_(c)gtK_(c)`, the net reaction will be in the backward direction. |
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| 1140. |
for the exothermic formation of sulphur trioxide from sulphur dioxide. And oxygen in the gas phase: `2SO_(2)(g) + O_(2)(g) hArr 2SO_(3) (g)` `K_(p) =40 .5 atm^(-1) at 900 K and Delta H =- 198 kJ` (i) Write the expression for the equilibrium constant for the reaction. (ii) At room temperature `(~~300 K)` will `K_(p)` be greater than less than or equal to `K_(p)` at 900 K . (iii) How will the equilibrium be affected if the volume of the vessel contaning the three gases is reduced, keeping the termperature constant ? What happens ? (iv) What is the effect of adding 1 mole of He (g) to a flask containing `SO_(2),O_(2)` and `SO_(3)` at equilibrium at constant temperatrue ? |
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Answer» (i) The expression for the equilibrium constant `(K_(p))` for the reaction is : `K_(p) =(P_(SO_3)^(2)(g))/(P_(SO_2)^(2)(g)xxP_(O_(2)))=40.5 atm^(-1)` (ii) the forward reaction is exothermic and the backward reaction is endothermic in nature. Therefore ,the increase in temperature favour the backward reaction. This means that the `K_(p)` at 300 K will be greater than the value at 900K. |
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| 1141. |
Which of the following is a buffer solution?A. `CH_(3)COOH +CH_(3)COONa`B. `CH_(3)COOH + CH_(3)COONH_(4)`C. `CH_(3)COOH + NH_(4)CI`D. `NaOH + HCI` |
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Answer» Correct Answer - A It is an acidic buffer |
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| 1142. |
For reacrtion `2NOC1(g) hArr 2NO(g) + C1_(2)(g)` , `K_(c)` at `427^(@)C` is `3xx10^(-6)` L `mol^(-1)`. The value of `K_(p)` is nearlyA. `7.5 xx 10^(-5)`B. `2.50 xx 10^(-5)`C. `2.5 xx 10^(-4)`D. `1.75 xx 10^(-4)` |
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Answer» Correct Answer - D `2 NOCI(g) hArr 2 NO(g) + CI_(2) (g)` `K_(p) =K_(c) (RT)^(Deltan) =3xx 10^(-6) (0.0821 xx 700)` `=1.72 xx 10^(-4)` |
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| 1143. |
Predict the acidic, basic or neutral nature of the solutions of the following salts : `NaCl, KBr, NaCN, NH_(4)NO_(3), NaNO_(2), KF`. |
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Answer» `NaCN, NaNO_(2), KF` solutions are basic, as they are salts of strong base, weak acid. NaCl, KBr solutions are neutral, as they are salts of strong acid, strong base. `NH_(4)NH_(3)` solution is acidic, as it is a salt of strong acid, weak base. |
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| 1144. |
Predict if the solutions of the following salts are neutral, acidic or basic: `NaCl, KBr, NaCN, NH_(4)NO_(3), NaNO_(2)` and `KF` |
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Answer» Correct Answer - `NaCl, KBr` solutions are neutral, NaCN, `NaNO_(2)` and `KF` solution are basic and `NH_(4)NO_(3)` solution is acidic. (i) NaCl: `NaCI+H_(2)Oleftrightarrowunderset("Strong base")(NaOH)+underset("Strong acid")(HCI)` Therefore, it is a neutral solution. (ii) KBr: `KBr+H_(2)Oleftrightarrowunderset("Strong base")(KOH)+underset("Strong acid")(HBr)` Therefore, it is a neutral solution. (iii) NaCN: `NaCN+H_(2)Oleftrightarrowunderset("Weak acid")(HCN)+underset("Strong base")(NaOH)` Therefore, it is a basic solution. (iv) `NH_(4)NO_(3)` `NH_(4)NO_(3)+H_(2)Oleftrightarrowunderset("Weak base")(NH_(4)OH)+underset("Strong acid")(HNO_(3))` Therefore, it is an acidic solution. (v) `NaNO_(2)` `NaNO_(2)+H_(2)Oleftrightarrowunderset("Strong base")( NaOH)+underset("Weak acid")(HNO_(2))` Therefore, it is a basic solution. (vi) KF `KF+H_(2)Oleftrightarrowunderset("Strong base")(KOH)+underset("Weak acid")(HF)` Therefore, it is a basic solution. |
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| 1145. |
The `K_(sp)` of CuS, `Ag_(2)S` and `HgS` are `10^(-31), 10^(-44)` and `10^(-54)` respectively. The solubility of these sulphides are in the order.A. `Ag_(2)S gt CuS lt HgS`B. `Ag_(2)S gt HgS gt CuS`C. `HgS gt Ag_(2)S gt CuS`D. `CuS gt Ag_(2)S gt HgS` |
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Answer» Correct Answer - D Higher the `K_(sp)` value more will be the solubility of the salt. |
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| 1146. |
A student prepared solutions of NaCl, `Na_(2)CO_(3) and NH_(4)Cl`. He put them separately in three test tubes. He forgot to label them. All solutions were colourless . How should he proceed to know the solutions present in the three test tubes ? |
| Answer» He can test with blue and red litmus solutions. NaCl solution is neutral. It will neither turn blue litmus red nor red litmus blue. `NH_(4)Cl` solution is acidic. It will turn blue litmus red but will have no effect on red litmus `Na_(2)CO_(3)` solution is basic. It will turn red litmus blue but will have no effect on blue litmus . | |
| 1147. |
Which of the following is a nonelectrolyte ?A. UreaB. MethanolC. EthanolD. All of these |
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Answer» Correct Answer - D Urea `[(NH_(2))_(2)CO]`, methanol `(CH_(3)OH)`, and ethanol `(C_(2)H_(5)OH)` are all nonelectrolytes because their aqueous solutions do not conduct electricity. |
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| 1148. |
20.0 L of 0.2 M weak acid `(pK_(a)=5.0)` is titrated against 0.2 M strong base. What is the pH at the equivalence point ?A. `5.0`B. `7.0`C. `9.0`D. `11.0` |
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Answer» Correct Answer - C At the equivalence point, solution will contain salt of weak acid and strong base. Its solution on hydrolysis will be basic with `pH gt 7`. Value of 9 may be chosen as the base is strong. |
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| 1149. |
Assertion: Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate `AgBr` rather than `AgCl`. Reason : `K_(sp)` of `AgClltK_(sp)` of `AgBr`.A. Statements -1 is true , statement -2 is also true, statement -2 is the correct explanation of statement -5B. Statement -1 is true , statement-2 is also true, statement-2 is not the correct explanation of statement-5C. Statement -1 is true, statement -2 is false.D. Statement -1 is false, satement-2 is true. |
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Answer» Correct Answer - C Correct statement-2 The `K_(sp)` of AgCI is more than that of AgBr. |
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| 1150. |
Predict if there will be any precipitate by mixing 50 mL of 0.01 M NaCl and 50 mL of M `AgNO_(3)` solution. The solubility product of AgCl is `1.5xx10^(-10)`.A. Since ionic product is greater than solubility product no precipitate will be formed.B. Since ionic product is lesser than solubility product, precipitation will occur .C. Since ionic product is greater than solubility product, precipitation will occur.D. Since ionic product and solubility product are same, precipitation will not occur. |
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Answer» Correct Answer - C `NaCl+AgNO_(3)toAgCl+NaNO_(3)K_(sp)=1.5xx10^(-10)` `[Ag^(+)]=(1)/(2)xx10^(-2)M=0.5xx10^(-2)M` `[Cl^(-)]=(1)/(2)xx10^(-2)M=0.5xx10^(-2)M` `K_(ip)=[Ag^(+)][Cl^(-)]=(0.5xx10^(-2))xx(0.5xx10^(-2))=2.5xx10^(-5)K_(ip)gtK_(sp)` When `K_(ip)gtK_(sp)`, it results in precipitation. |
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