MCQ 511 Mark
For the reactipon, $\text{SO}_2(\text{g})+\frac{1}{2}\text{O}_2\text{(g)}\rightleftharpoons\text{SO}_3(\text{g})$ if $K_p=K_c(R T)^X$ where, the symbols have usual meaning, then the value of $x$ is $($assuming ideality$)$
- A
$-1$
- ✓
$-\frac{1}{2}$
- C
$\frac{1}{2}$
- D
$1$
AnswerCorrect option: B. $-\frac{1}{2}$
View full question & answer→MCQ 521 Mark
$\ce{ {SO}_2 + {O}_2 \rightleftharpoons 2 {SO}_3} +$ Heat The equilibrium reaction proceeds in forward direction by:
- ✓
Addition of $O_2$
- B
Removal of $O_2$
- C
- D
AnswerCorrect option: A. Addition of $O_2$
According to Le$-$Chatelier's principle, equilibrium shift towards forward direction by addition of reactant.
View full question & answer→MCQ 531 Mark
Equal volume of following $Ca^{2+}$ and $F$ solution are mixed. In which of the Solutions will precipitation occur? $[K_{SP}$ of $\ce{CaF_2} = 1.7 × 10^{-10}$
- A
$10^{-2}\text{M Ca}^{2+}+10^{-5}\text{MF}^-$
- B
$10^{-3}\text{M Ca}^{2+}$ and $10^{-3}\text{MF}^-$
- ✓
$10^{-2}\text{M Ca}^{2+}+10^{-3}\text{MF}^-$
- D
$10^{-3}\text{M Ca}^{2+}$ and $10^{-5}\text{MF}^-$
AnswerCorrect option: C. $10^{-2}\text{M Ca}^{2+}+10^{-3}\text{MF}^-$
$\because\text{I.P}=(\text{Ca}^{2+})(\text{F}^-)^2$
$=10^{-2}\times(10^{-3})^2=10^{-8}>\text{K}_{\text{SP}}$
$\therefore$ precipitation will occur.
View full question & answer→MCQ 541 Mark
What will happen when $\ce{CH_3COONa}$ is added to an aqueous solution of $\ce{CH_3COOH}$?
- A
The $pH$ of the solution decreases.
- ✓
The $pH$ of the solution increases.
- C
The $pH$ of the solution remains unaltered.
- D
An acidic salt is produced.
AnswerCorrect option: B. The $pH$ of the solution increases.
$\text{CH}_3\text{COOH}⇋\text{CH}_3\text{COO}^−+\text{H}^+$
If $\ce{CH_3COONa}$ is added, equilibrium will shift backward due to the common ion effect. $H^+$ will decrease.
View full question & answer→MCQ 551 Mark
In the equilibrium, $\text{AB}\rightleftharpoons\text{A}+\text{B}$ if the equilibrium concentration of $A$ is double, then equilibrium concentration of $B$ will be:
- ✓
- B
- C
$\frac{1}{4}^{th}$
- D
$\frac{1}{8}^{th}$
Answer$\text{AB}\rightleftharpoons\text{A}+\text{B}$
or $\text{K}=\frac{[\text{A}][\text{B}]}{[\text{AB}]}$
If concentration of $A$ is doubled, the equilibrium concentration of $B$ becomes half to maintain $K$ constant.
View full question & answer→MCQ 561 Mark
In the melting of ice, which one of the conditions will be more favourable?
- ✓
High temperature and high pressure.
- B
Low temperature and low pressure.
- C
Low temperature and high pressure.
- D
High temperature and low pressure.
AnswerCorrect option: A. High temperature and high pressure.
Since ice melts with the absorption of heat and decreases in volume, hence both temperature and pressure affect the melting of ice. Since the change of ice into water is an endothermic process hence with rising of temperature ice melts into water. Since the volume of ice is more than that of water so an increase of pressure favour melting.
View full question & answer→MCQ 571 Mark
Which of the following is correct regarding buffer sol?
AnswerCorrect option: C. It shows little change in $pH$ on adding small amount of acid or base.
A buffer solution shows little change in $pH$ on adding small amount of acid or base.
View full question & answer→MCQ 581 Mark
The addition of $\text{NaCl}$ to $\text{AgCl}$ decreases the solubility of $\text{AgCl}$ because:
- A
Solubility product decreases.
- ✓
Solubility product remains constant.
- C
Solution becomes unsaturated.
- D
Solution becomes supersaturated.
AnswerCorrect option: B. Solubility product remains constant.
$[Cl^-]$ increases but $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right][\mathrm{Cl}]$remains constant. So, $[Ag^+]$ decreases.
View full question & answer→MCQ 591 Mark
Amongst the following hydroxides, the one which has the highest value of $\text{Ksp}$ at $25^\circ C$ is:
- A
$\text{KOH}$
- ✓
$\text{CsOH}$
- C
$\text{LiOH}$
- D
$\text{RbOH}$
AnswerCorrect option: B. $\text{CsOH}$
Going down Group$-I$ in the periodic nature, the ionic nature of the hydroxides increases. As ionic nature increases, the solubility also increases.
View full question & answer→MCQ 601 Mark
In the formation of nitric acid $N_2$ and $O_2$ are made to combine. Thus,$N_2 + O_2⇌ 2NO−$Heat which of the following condition will favour the formation of $NO$?
View full question & answer→MCQ 611 Mark
The equilibrium $\ce{2SO_2(g) + O_2(g) ⇌ 2SO_3(g)}$ shifts forward if:
AnswerCorrect option: B. An adsorbent is used to remove $SO_3$ as soon as it is formed.
Removal of any product or adding of reactants favours forward reaction i.e.,$ SO_3$ formation. This is according to Le$-$ Chatelier's principle.
View full question & answer→MCQ 621 Mark
The rate of the reaction $\ce{2NO + Cl_2 \rightarrow 2NOCls}$ given by the rate equation. rate $=\ce{ k[NO]^2[Cl_2]}$ The value of the rate constant can be increased by:
AnswerCorrect option: C. Increasing the temperature.
Rate constant is only affected by the temperature, it does not get affected by the increase in concentration of $NO$ and $\ce{Cl_2}$ .
View full question & answer→MCQ 631 Mark
When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. On the basis of this information mark the correct answer.
$[\text{Co(H}_2\text{O})_6]^{3+}\text{(aq)}+4\text{Cl}^-\text{(aq)}\rightleftharpoons[\text{CoCl}_4]^{2-}\text{(aq)}+6\text{H}_2\text{O (l)}\\ \ \ \ \ \ \ \ ^\text{(pink)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(blue)}$
- ✓
$\Delta\text{H}>0$ for the reaction.
- B
$\Delta\text{H}<0$ for the reaction.
- C
$\Delta\text{H}=0$ for the reaction.
- D
The sign of $\Delta\text{H}$ cannot be predicted on the basis of this information.
AnswerCorrect option: A. $\Delta\text{H}>0$ for the reaction.
On cooling the mixture reaction moves towards backwards direction it means it is an endothermic reaction i.e., $\Delta\text{H}>0.$
View full question & answer→MCQ 641 Mark
Which of the following has maximum $pK_a$:
- A
$ \mathrm{CH}_2 \mathrm{FCOOH}$
- B
$\mathrm{CH}_3 \mathrm{ClCOOH} $
- ✓
$\mathrm{CH}_3 \mathrm{COOH} $
- D
$ \mathrm{HCOOH} $
AnswerCorrect option: C. $\mathrm{CH}_3 \mathrm{COOH} $
View full question & answer→MCQ 651 Mark
In the following reaction, $\begin{matrix}\text{H}_2\text{O}(\text{l})&+&\text{H}_2\text{O}(\text{l})&\rightleftharpoons&\text{H}_3\text{O}^+(\text{aq})&+&\text{OH}^-(\text{aq})\\\text{Acid}&&\text{ X}&&\text{Conjugate Acid}&&\text{Y}\end{matrix} X$ and $Y$ respectively are:
View full question & answer→MCQ 661 Mark
$\text{A}+\text{B}\rightleftharpoons\text{C}+\text{D}.$ If the concentration of $A$ and $B$ are equal at equilibrium and concentration of $D$ will be twice that of $A$, then what will be the equilibrium constant of the reaction?
- ✓
$4$
- B
$6$
- C
$\frac{4}{5}$
- D
$\frac{6}{7}$
View full question & answer→MCQ 671 Mark
Which of the following options will be correct for the stage of half completion of the reaction $\text{A}\rightleftharpoons\text{B}.$
- ✓
$\Delta\text{G}^\ominus=0$
- B
$\Delta\text{G}^\ominus>0$
- C
$\Delta\text{G}^\ominus<0$
- D
$\Delta\text{G}^\ominus=-\text{RT}$ ln $2$
AnswerCorrect option: A. $\Delta\text{G}^\ominus=0$
$\text{A}\rightleftharpoons\text{B}$
$\Delta\text{G}^\ominus=-\text{RT}$ ln $K$
At the stage of half completion of reaction $[A] = [B],$
Therefore, $K = 1$. Thus, $\Delta\text{G}^\ominus=0$
View full question & answer→MCQ 681 Mark
Strong acid dissociates completely in water, the resulting base formed would be very weak. The reason is that:
- A
Strong acids have strong conjugate bases.
- B
Strong acids have strong conjugate acids.
- ✓
Strong acids have very weak conjugate bases.
- D
Strong acids have very weak conjugate acids.
AnswerCorrect option: C. Strong acids have very weak conjugate bases.
View full question & answer→MCQ 691 Mark
In the equilibrium reaction, $\mathrm{N}_2+3 \mathrm{H}_2 \Leftrightarrow 2 \mathrm{NH}_3$, the sign of $\triangle H$ accompanying the reaction is:
- A
- ✓
- C
May be positive or negative.
- D
AnswerThe reaction for the formation of ammonia is $\mathrm{N}_2+3 \mathrm{H}_2 \Leftrightarrow 2 \mathrm{NH}_3+92.2 \mathrm{~kJ}$.
View full question & answer→MCQ 701 Mark
Acidity of $BF_3$ can be explained on the basis of which of the following concepts?
- A
- B
- ✓
- D
Bronsted Lowry as well as Lewis concept.
AnswerAccording to Lewis concept, a positively charged or an electron deficient species acts as Lewis acid. $BF_3$ is an electron deficient compound with B having 6 electrons only.
View full question & answer→MCQ 711 Mark
Which one does not give a buffer solution?
- ✓
Ammonia and sodium hydroxide in water.
- B
Sodium acetate and acetic acid in water.
- C
Ammonia and ammonium chloride in water.
- D
Sodium acetate and hydrochloric acid in water.
AnswerCorrect option: A. Ammonia and sodium hydroxide in water.
Buffer solution is a mixture of weak acid and its conjugate base or weak base and its conjugate acid. So ammonia $($strong base$)$ and sodium hydroxide does not give a buffer.
View full question & answer→MCQ 721 Mark
The addition of $\text{NaCl}$ to $\text{AgCl}$ decreases the solubility of $\text{AgCl}$ because $.........$
- A
Solubility product decreases.
- B
Solubility product remains constant.
- C
Solution becomes unsaturated.
- ✓
Solution becomes super saturated.
AnswerCorrect option: D. Solution becomes super saturated.
$\text{NaCl}$ is highly soluble and when it is added to $\text{AgCl}$ it decreases the solubility of $\text{AgCl}$ because of common ion $\ce{Cl^-}$ and solution become super saturated.
View full question & answer→MCQ 731 Mark
Which of the following will produce a buffer solution when mixed in equal volumes?
- A
$0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_4 \mathrm{OH}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$.
- B
$0.05 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_4 \mathrm{OH}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$.
- ✓
$0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_4 \mathrm{OH}$ and $0.05 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$.
- D
$0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{CH}_4 \mathrm{COONa}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NaOH}$.
AnswerCorrect option: C. $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_4 \mathrm{OH}$ and $0.05 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$.
In $(c)$, all $\text{HCl}$ will be neutralized and $\mathrm{NH}_4 \mathrm{Cl}$ will be formed. Also some $\mathrm{NH}_4 \mathrm{OH}$ will be left unneutralized. Thus, the final solution will contain $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{NH}_4 \mathrm{Cl}$ and hence will form a buffer.
View full question & answer→MCQ 741 Mark
$\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_3 \mathrm{COOH}$ is $1.8 \times 10^{-5}$ and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 \times 10^{-5}$ . The $pH$ of ammonium acetate will be
- A
$7.005$
- B
$4.75$
- ✓
$7.0$
- D
Between $6$ and $7$
AnswerAmmonium acetate is a salt of weak acid and weak base.
$\text{pH}=\frac{1}{2}\big[\text{pK}_\text{w}+\text{pK}_\text{a}+\text{pK}_\text{b}\big]$
$=\frac{1}{2}\big[14-\log(1.8\times10^{-5})+\log(1.8\times10^{-5})\big]=7.0$
View full question & answer→MCQ 751 Mark
A solution of an acid has $pH = 4.70.$ Find out the concentration of $OH^-$ ions $(pK_w = 14).$
- ✓
$ 5 \times 10^{-10} $
- B
$ 6 \times 10^{-10} $
- C
$ 2 \times 10^{-5} $
- D
$ 9 \times 10^{-5} $
AnswerCorrect option: A. $ 5 \times 10^{-10} $
View full question & answer→MCQ 761 Mark
Given the chemical equilibrium,$ A ⇌ B + C$, where $\triangle H_{rxn}$ is negative, what effect increasing the temperature $($at constant pressure$)$ would have on the system at equilibrium?
Answer$A \rightarrow B + C\ ; \triangle H < 0$
Here, forward reaction is exothermic. So, if temperature is increased then as per Le Chatelier's principle, the equilibrium shifts to left side $($backward direction$)$.
View full question & answer→MCQ 771 Mark
Which of the following statements is incorrect?
- A
In equilibrium mixture of ice and water kept in perfectly insulated flask mass of ice and water does not change with time.
- ✓
The intensity of red colour increases when oxalic acid is added to a solution containing iron $(III)$ nitrate and potassium thiocyanate.
- C
On addition of catalyst the equilibrium constant value is not affected.
- D
Equilibrium constant for a reaction with negative $\Delta\text{H}$ value decreases as the temperature increases.
AnswerCorrect option: B. The intensity of red colour increases when oxalic acid is added to a solution containing iron $(III)$ nitrate and potassium thiocyanate.
$\text{Fe}^{3+}+\text{SCN}^-\rightleftharpoons\text{FeSCN}^{2+}\text{(Red)}.$
When oxalic acid is added to a solution containing iron nitrate and potassium thiocyanate, oxalic acid reacts with $\text{Fe}{3+}$ ions to form a stable complex ion $[\text{Fe}\text{(C}_2\text{O}_4)3]^{3-},$ thus, decreasing the concentration of free $\text{Fe}^{3+}$ ions which in mm decreases the intensity of red colour.
$\text{Fe}^{3+}+\text{SCN}^-\rightleftharpoons\text{[Fe(SCN)}]^{2+}\text{(Red)}.$
View full question & answer→MCQ 781 Mark
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g}),$ if the degree of dissociation is a at equilibrium pressure$ 'p',$ then the equilibrium constant for the reaction is:
- A
$\text{K}_{\text{p}}=\frac{\alpha^2}{1+\alpha^2\text{P}}$
- B
$\text{K}_{\text{p}}=\frac{\alpha^2\text{P}^2}{1-\alpha^2}$
- C
$\text{K}_{\text{p}}=\frac{\alpha\text{P}^2}{1-\alpha^2}$
- ✓
$\text{K}_{\text{p}}=\frac{\alpha^2\text{P}}{1-\alpha^2}$
AnswerCorrect option: D. $\text{K}_{\text{p}}=\frac{\alpha^2\text{P}}{1-\alpha^2}$
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2(\text{g})$
$\begin{matrix}\text{Initial}&1&0&0\\\text{At equilibrium}&1-\alpha&\alpha&\alpha\end{matrix}$
Total no. of moles $=1-\alpha+\alpha+\alpha=1+\alpha$
$\text{pPCl}_5=\frac{1-\alpha}{1+\alpha}$
$\text{pCl}_2=\text{pCl}_3=\frac{\alpha}{1+\alpha}$
$\text{K}_{\text{p}}=\frac{(\text{PCl}_3)(\text{Cl}_2)}{(\text{PCl}^-_3)}$
$=\frac{\frac{\alpha}{1+\alpha}\times\frac{\alpha}{1+\alpha}\times\text{p}^2}{\frac{1-\alpha}{1+\alpha}\times\text{p}}$
$=\frac{\alpha^2}{1-\alpha^2}\times\text{p}$
View full question & answer→MCQ 791 Mark
In a reversible reaction $\mathrm{H}_2+\mathrm{I}_2 \leftrightharpoons 2 \mathrm{HI}$, if the concentration of $H_2$ and $I_2$ are increased, the value of $Kc:$
AnswerThe magnitude of the equilibrium constant is not affected by the changes in concentrations of reactants and products, pressure and volume. Thus, when the concentrations of hydrogen and iodine are increased, the value of the equilibrium constant remains the same.
View full question & answer→MCQ 801 Mark
In the presence of a common ion $($incapable of forming complex ion$)$, the solubility of salt $.........$ in solution.
Answer$ \mathrm{AB} \rightarrow \mathrm{A}^{+}+\mathrm{B}^{-} $
$ \mathrm{BC} \rightarrow \mathrm{B}^{+}+\mathrm{C}^{-} $
Since $B^+$ is incapable of forming a complex salt it tends to decrease the solubility by $Le-$Chatelier's principle.
View full question & answer→MCQ 811 Mark
Degree of ionisation does not depend on:
- A
- B
Ature of the electrolyte.
- C
- ✓
Molecular mass of the electrolyte.
AnswerCorrect option: D. Molecular mass of the electrolyte.
Degree of ionization$(\alpha)$ depends on$-$
$(1)$ Concentration of solute.
$(2)$ Temperature.
$(3)$ Nature of electrolysis.
$(4)$ Nature of solvent.
$(5)$ Dilution.
$"α"$ does not depend on molecular mass of electrolyte.
View full question & answer→MCQ 821 Mark
Strong electrolyte of the following is?
- A
$\ce{01M HAc}$
- ✓
$\ce{0.1M HCl}$
- C
$\ce{0.1M KCl}$
- D
$\ce{0.1M NaCl}$
AnswerCorrect option: B. $\ce{0.1M HCl}$
View full question & answer→MCQ 831 Mark
Calculate the molar solubility $(S)$ of a salt like zirconium phosphate of molecular formula $(\text{Zr}^{4+})_3(\text{PO}^{3-}_4)_4.$
- A
$\Big(\frac{\text{K}_{\text{sp}}}{9612}\Big)^{\frac{1}{8}}$
- ✓
$\Big(\frac{\text{K}_{\text{sp}}}{6912}\Big)^{\frac{1}{7}}$
- C
$\Big(\frac{\text{K}_{\text{sp}}}{5348}\Big)^{\frac{1}{6}}$
- D
$\Big(\frac{\text{K}_{\text{sp}}}{8435}\Big)^{\frac{1}{7}}$
AnswerCorrect option: B. $\Big(\frac{\text{K}_{\text{sp}}}{6912}\Big)^{\frac{1}{7}}$
$[\text{Zr}^{4+}]=3\text{S}$ and $\text{PO}^{3-}_4]=4\text{S}$
and $\text{K}_{\text{sp}}=(3\text{S})^3(4\text{S})^4=6912(\text{S})^7$
or $\text{S}=\Big\{\ \frac{\text{K}_{\text{sp}}}{(3^3\times4^4)}\Big\}^{\frac{1}{7}}$
$=\Big(\frac{\text{K}_{\text{sp}}}{6912}\Big)^{\frac{1}{7}}$
View full question & answer→MCQ 841 Mark
In the reaction,$\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})-180.7 \mathrm{~kJ}$, on increasing the temperature, the production of $NO$ :
AnswerGiven $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})-180.7 \mathrm{~kJ}$
For this reaction, $\ce{\triangle H = +180.7 kJ}$
Positive value of $\triangle H$ shows that the reaction is endothermic.
According to Le$-$Chatelier's principle, in case of endothermic reaction, increase in temperature, shifts the reaction towards product side.
View full question & answer→MCQ 851 Mark
Predict which of the following reaction will have appreciable concentration of reactants and products?
- A
$\text{Cl}_2(\text{g})\rightleftharpoons2\text{Cl(g)};\text{K}_{\text{c}}=5\times10^{-39}$
- B
$\text{Cl}_2(\text{g})+\text{2NO}(\text{g})\rightleftharpoons2\text{NOCl}(\text{g});\text{K}_{\text{c}}=37\times10^8$
- ✓
$\text{Cl}_2\text{(g)}+2\text{NO}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g});\text{K}_{\text{c}}=1.8$
- D
All have appreciable concentration of reactants and products.
AnswerCorrect option: C. $\text{Cl}_2\text{(g)}+2\text{NO}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g});\text{K}_{\text{c}}=1.8$
View full question & answer→MCQ 861 Mark
The following reaction goes to completion in lime kiln because : $\mathrm{CaCO}_3 \rightleftharpoons \mathrm{CaO}+\mathrm{CO}_2(\mathrm{~g})$
- A
- B
$\ce{CaO}$ is more stable than $\ce{CaCO_3}$.
- C
$\ce{CaO}$ is not dissociated.
- ✓
$\ce{CO_2}$ escapes continuously.
AnswerCorrect option: D. $\ce{CO_2}$ escapes continuously.
In lime kilns, $\ce{CO_2}$ formed continues to escape into the atmosphere and equilibrium is never established.
View full question & answer→MCQ 871 Mark
Dissociation events in $\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}$ is termed as ionization because:
- ✓
The electron is initially shared between both atoms, thus the dissociation event into ions involves the transfer of an electron from one atom to the other.
- B
It does not involve any electron transfer.
- C
- D
No charge species formed.
AnswerCorrect option: A. The electron is initially shared between both atoms, thus the dissociation event into ions involves the transfer of an electron from one atom to the other.
$\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}$
This dissociation is termed as ionization because initially there is a polar covalent compound $\mathrm{H}_2 \mathrm{O}$.
This dissociation leads to the formation of ions $H^+$ and $OH^-$
This dissociation involves a transfer of an electron from one atom to another leading to the formation of charged species.
View full question & answer→MCQ 881 Mark
A buffer solution is a solution whose $\ce{pH}$ value on keeping in the air :
- A
- B
- C
May increase or decrease.
- ✓
View full question & answer→MCQ 891 Mark
The ionic product of water $ .......... $ if a few drops of acid or base are added to it.
AnswerThe ionic product of water at a particular temperature is constant and has no effect of acid or base addition.
View full question & answer→MCQ 901 Mark
Which of the following species is amphoteric in nature.
- A
$\text{H}_3\text{O}^+$
- ✓
$\text{Cl}^-$
- C
$\text{HSO}^-_4$
- D
$\text{CO}^{2-}_3$
AnswerCorrect option: B. $\text{Cl}^-$
$\text{HSO}^-_3$ because if can gain $\ce{H}^+$ as well as lose $\ce{H}^+$.
View full question & answer→MCQ 911 Mark
The solubility product $K_{sp}$ of the sparingly soluble salt $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $4 \times 10^{-12}$. The molar solubility of the salt is :
- ✓
$1.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} $
- B
$ 2 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} $
- C
$ 1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} $
- D
$ 2 \times 10^{-12} \mathrm{~mol} \mathrm{~L}^{-1} $
AnswerCorrect option: A. $1.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} $
View full question & answer→MCQ 921 Mark
Which of the following factors will favour the reverse reaction in a chemical equilibrium?
- A
Increase in concentration of one of the reactants.
- ✓
Increase in concentration of one of the products.
- C
Removal of one of the products regularly.
- D
AnswerCorrect option: B. Increase in concentration of one of the products.
$\ce{A + B \rightleftharpoons C + D}$
According to Le ChateIier's principle, as the concentration of the products are increased, the reaction proceeds in the backward direction.
View full question & answer→MCQ 931 Mark
$\mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_5(\mathrm{~g})+$ energy For the reaction above, what is the effect of increasing the pressure?
- ✓
Increased production of products.
- B
Wild fluctuations in the amounts of reactants and products.
- C
Increased production of reactants.
- D
No impact on the equilibrium.
AnswerCorrect option: A. Increased production of products.
$\mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_5(\mathrm{~g})+$ energy
In this reaction, the forward reaction is accompanied by a decrease of moles of gaseous species. So, if pressure on the system is increased then, as per Le Chatelier's principle, the equation shifts in direction in which a decrease in total no. of moles takes place i.e. in formation of products in this case.
View full question & answer→MCQ 941 Mark
Addition of $\text{HCl}$ will not suppress the ionization of :
AnswerAny acid weaker than $\text{HCl}$ will be suppressed by $\text{HCl}$. Hence, among the given options, only sulphuric acid is an acid with comparable acidity strength to $\text{HCl}$. The same can also be verified using $\text{Ka}$ values from the data.
View full question & answer→MCQ 951 Mark
The $\ce{pH}$ value of blood does not appreciably change by a small addition of an acid or a base, because the blood :
- A
- B
Can be easily coagulated.
- C
Contains iron as a part of the molecule.
- ✓
Contains serum protein which acts as buffer.
AnswerCorrect option: D. Contains serum protein which acts as buffer.
The buffer system present in serum is $\mathrm{H}_2 \mathrm{CO}_3+\mathrm{NaHCO}_3$ and as we know that a buffer solution resist the change in $\ce{pH}$ therefore $\ce{pH}$ value of blood does not change by a small addition of an acid or a base.
View full question & answer→MCQ 961 Mark
A solution which maintains constant $\ce{pH}$ when small amounts of acid or alkali are added is known as $ .........$
AnswerA buffer solution is one which resists changes in $\ce{pH}$ when small quantities of an acid or an alkali are added to it.
View full question & answer→MCQ 971 Mark
If the temperature of the system of equilibrium is increased, the equilibrium will shift in the direction which $ ......... $ heat.
AnswerIf a chemical system at equilibrium experiences a change in concentration, temperature, volume or pressure then, the equilibrium shifts to counteract the imposed change and a new equilibrium is established.
Therefore, If the temperature of the system of equilibrium is increased, the equilibrium will shift in the direction which absorbs heat. It is a direct implication of the Le Chatelier's principle $($effect of pressure and temperature$).$
View full question & answer→MCQ 981 Mark
For the reaction $\text{H}_2\text{(g)}+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI(g)},$ the standard free energy is $\Delta\text{G}^\ominus > 0.$ The equilibrium constant $(K)$ would be $ .........$
- A
$K = 0.$
- B
$K > 1.$
- C
$K = 1.$
- ✓
$K < 1.$
AnswerCorrect option: D. $K < 1.$
$\Delta\text{G}^\circ=-\text{R}\text{T }1\text{nK}$
If $\Delta\text{G}^\ominus > 0,$ then $-\Delta\text{G}^\ominus/\text{RT}$ is negative, and $\text{e}^{\Delta\text{G}^\ominus/\text{RT}} < 1.$ That is $K < 1,$ which implies a non $-$ spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.
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$2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2$ The equilibrium constant of the above reaction is $6.4$ at $300K$. If $0.25$ mole each of $\ce{H_2}$ and $\ce{I_2}$ are added to the system, the equilibrium constant will be :
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If $\text{HCl}$ is added to pure water at $25^\circ C$ the ionic product of water will be :
- A
$ > 10^{-14} $
- B
$ < 10^{-14} $
- ✓
$ 10^{-14} $
- D
$ > 10^{-10} $
AnswerCorrect option: C. $ 10^{-14} $
For ionic product of pure water :
$ {\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{M}} $
$ {\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{M}} $
$\text { Product, } \mathrm{K}_{\mathrm{W}}=10^{-14}$
lonic Product $\text{Kw}$ of water is dependent only on temperature so it will remain same,
$ \mathrm{K}_{\mathrm{W}}=10^{-14}$.
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