MCQ 11 Mark
Which of the following is the example of a reversible reaction?
- A
$\text{Pb(NO}_3)_2(\text{aq})+2\text{NaI(aq)}\xrightarrow{\ \ \ \ \ \ \ \ }\text{PbI}_2(\text{s})+2\text{NaNO}_3(\text{aq})$
- B
$\text{2Na(s)}+2\text{H}_2\text{O}(\text{l})\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{2NaOH}(\text{aq})+\text{H}_2(\text{g})$
- C
$\text{AgNO}_3(\text{aq})+\text{HCl}(\text{aq})\xrightarrow{\ \ \ \ \ \ \ \ }\text{AgCl}(\text{s})+\text{HNo}_3(\text{aq})$
- ✓
$\text{KNO}_3(\text{aq})+\text{NaCl(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{KCl(aq)}+\text{NaNO}_3\text{(aq)}$
AnswerCorrect option: D. $\text{KNO}_3(\text{aq})+\text{NaCl(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{KCl(aq)}+\text{NaNO}_3\text{(aq)}$
View full question & answer→MCQ 21 Mark
What is $\ce{pH}$ of resulting solution when equal volume when equal of $0.1M ,\ce{NaOH}$ and $0.01M, \ce{HCl}$ are mixed? $[\log 4.5 = 0.65]$
- A
$7$
- B
$1.04$
- ✓
$12.65$
- D
$2.0$
AnswerCorrect option: C. $12.65$
Number of mole of $[\text{OH}^-]=\frac{0.1}{2}=0.05$ on mixing
$[\text{H}^+]=\frac{0.01}{2}=0.005$
No. of mole of $\ce{[OH}^-] $ left $= 0.05 - 0.005 = 0.045$
$\text{pOH} = -\log 4.5 \times 10-2$
$= -0.65 + 2.00 = 1.35$
$\ce{pH } = 14 - 135 = 12.65$
View full question & answer→MCQ 31 Mark
The $\ce{pH}$ of neutral water at $25^\circ C$ is $7.0.$ As the temperature increases, ionisation of water increases, however, the concentration of $\ce{H}^+$ ions and $\ce{OH}^-$ ions are equal. What will be the $\ce{pH}$ of pure water at $60^\circ C\ ?$
- A
Equal to $7.0$
- B
Greater than $7.0$
- ✓
Less than $7.0$
- D
AnswerCorrect option: C. Less than $7.0$
As $K_w$ increases $[\text{H}^+][\text{OH}^-] > 10^{-14}$
As $[\text{H}^+]=[\text{OH}^-]$
or $[\text{H}^+]^2=10^{-14}$
or $[\text{H}^+]>10^{-7}\text{M}$
$\ce{pH } < 7$
View full question & answer→MCQ 41 Mark
$2\text{NO}_2(\text{g})\rightleftharpoons\text{N}_2\text{O}_4(\text{g})+60.0\text{KJ,}$ the increase in temperature :
AnswerCorrect option: B. Favour the decomposition of $\ce{N_2O_4}$.
It will favour backward reaction because process is exothermic.
View full question & answer→MCQ 51 Mark
Which is not a buffer solution :
- A
$ \mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH} $
- B
$ \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa} $
- ✓
$ \mathrm{CH}_3 \mathrm{COONH}_4$
- D
$ \mathrm{NH}_4 \mathrm{NO}_3$
AnswerCorrect option: C. $ \mathrm{CH}_3 \mathrm{COONH}_4$
A buffer solution either is a mixture of a weak acid and its salt with strong base or a mixture of a weak base and its salt with strong acid.
Hence, clearly $ \mathrm{CH}_3 \mathrm{COONH}_4$
is not a buffer solution.
View full question & answer→MCQ 61 Mark
What will be the correct order of vapour pressure of water, acetone and ether at $30^\circ C$. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?
- A
Water $ < $ ether $ < $ acetone.
- ✓
Water $ < $ acetone $ < $ ether.
- C
Ether $ < $ acetone $ < $ water.
- D
Acetone $ < $ ether $ < $ water.
AnswerCorrect option: B. Water $ < $ acetone $ < $ ether.
Greater the boiling point, less is the vapour pressure.
Hence, the correct order of vapour pressures will be:
water $ < $ acetone $ < $ ether.
View full question & answer→MCQ 71 Mark
The solubility of $\ce{CO_2}$ in water increases with :
- A
- B
Reduction of gas pressure.
- ✓
Increase in gas pressure.
- D
AnswerCorrect option: C. Increase in gas pressure.
When carbon dioxide reacts with water it forms carbonic acid. Increasing the pressure of carbon dioxide makes the reaction feasible in the forward direction and hence solubility of $\ce{CO_2}$ increases.
View full question & answer→MCQ 81 Mark
$\ce{H_2O}$ is a:
- A
- B
- ✓
Both proton donor and proton acceptor.
- D
Neither proton donor, nor proton acceptor.
AnswerCorrect option: C. Both proton donor and proton acceptor.
A compound such as water $\ce{(H_2O)}$ has many interesting properties. Water molecules can accept a proton to act as Bronsted $-$ Lowry bases in certain circumstances. The following is an example of $\text{HCl}$ dissolving in water :
$\text{HCl}+\text{H}_2\text{O}(\ell)\rightarrow\text{H}_3\text{O}^+_\text{(aq)}+\text{C}1^-_\text{(aq)}$
The another possibility is water can act like a Bronsted $-$ Lowry acid by donating a proton. Water donates a proton to a proton $-$ accepting amide ion in the presence of ammonia, resulting in the following product:
$\text{H}_2\text{O}\ell{(ℓ)}+\text{NH}^-_2\text{(aq)}\rightarrow\text{OH}^-_{\text{(aq)}}+\text{NH}_3\text{(aq)}$
View full question & answer→MCQ 91 Mark
$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \leftrightharpoons 2 \mathrm{SO}_3(\mathrm{~g})+\mathrm{QkJ}$ In the above reaction, how can the yield of product be increased without increasing the pressure?
- A
By increasing temperature.
- ✓
By decreasing temperature.
- C
By increasing the volume of the reaction vessel.
- D
By the addition of the catalyst.
AnswerCorrect option: B. By decreasing temperature.
Since the reaction is exothermic in the forward direction, a decrease in temperature favours the forward reaction.
View full question & answer→MCQ 101 Mark
On increasing the pressure, in which direction will the gas phase reaction proceed to re $-$ establish equilibrium, is predicted by applying the Le Chatelier’s principle. Consider the reaction.
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)}$
Which of the following is correct, if the total pressure at which the equilibrium is established, is increased without changing the temperature?
AnswerCorrect option: A. $K$ will remain same.
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)}$
According to Le Chatelier’s principle, at constant temperature, the equilibrium composition will change but $K$ will remain same.
View full question & answer→MCQ 111 Mark
The solubility product of a sparingly soluble salt $AB$ at room temperature is $1.21 \times 10^{-6},$ its molar solubility is :
- A
$1.21 \times 100M.$
- B
$1.1 \times 10^{-4}M.$
- ✓
$1.1 \times 10^{-3}M.$
- D
AnswerCorrect option: C. $1.1 \times 10^{-3}M.$
$\text{S}=\sqrt{\text{K}_{\text{sp}}}$
$=\sqrt{1.21\times10^{-6}}$
$=1.1\times10^{-3}\text{M.}$
View full question & answer→MCQ 121 Mark
The solubility of $\text{AgI}$ in $\text{NaI}$ solutions is less than that in pure water because :
Answer$ \mathrm{AgI} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{I}^{-}$
$ \mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{I}^{-}$
Sodium iodide is a strong electrolyte and is completely dissociated.
This increases the iodide ion concentration in solution and suppresses the ionization of $\text{AgI}.$
View full question & answer→MCQ 131 Mark
Addition of water to this solution will not change $[\mathrm{H}_3 \mathrm{O}^{+}]$.
AnswerAddition of water to $X$ solution does not change the $\ce{pH}$ of the solution, which means the concentration of the species present changes in such a way that the $\ce{pH}$ remains the same. This is possible for acid/base buffer.
$\text{pH}=\text{pk}_\text{a}+\log\frac{[\text{acid}]}{[\text{salt}]}$
Since the volume changes the same for both salt and acid, the ratio $\frac{\text{[salt]}}{\text{[acid]}}$remains the same and hence $\ce{pH}$ also remains the same. Same in the case of basic buffer solution.
View full question & answer→MCQ 141 Mark
The concentration of hydrogen ion in a sample of soft drink is $3.8 \times 10^{-3}M$. What is its $\ce{pH}\ ?$
- A
$4.32$
- B
$5.12$
- C
$3.31$
- ✓
$2.42$
AnswerCorrect option: D. $2.42$
View full question & answer→MCQ 151 Mark
Buffer solution can be obtained by mixing aqueous solution of $ .........$
- A
Sodium acetate and excess of $\text{HCl}.$
- ✓
Sodium acetate and acetic acid.
- C
Sodium chloride and $\text{HCl}.$
- D
Acetic acid and excess of $\text{NaOH}.$
AnswerCorrect option: B. Sodium acetate and acetic acid.
A mixture of acetic acid $($a weak acid$)$ and sodium acetate $($its salt with strong base sodium hydroxide$)$ acts as acidic buffer.
Option $A, C$ and $D$ does not contain weak acid/ weak base and its salt thus it does not form buffer solution.
View full question & answer→MCQ 161 Mark
Which of the following is not a general characteristic of equilibria involving physical processes?
- A
Equilibrium is possible only in a closed system at a given temperature.
- B
All measurable properties of the system remain constant.
- ✓
All the physical processes stop at equilibrium.
- D
The opposing processes occur at the same rate and there is dynamic but stable condition.
AnswerCorrect option: C. All the physical processes stop at equilibrium.
All the physical processes like melting of ice and freezing of water, etc. do not stop at equilibrium.
View full question & answer→MCQ 171 Mark
What will be $\ce{pH}$ of $0.01M , \ce{CH_3COOH}\ ? (K_a = 1.80 \times 10^{-5})$
Answer$\text{pH}=-\log(\text{H}^+)$
$=-\log\sqrt{\text{K}_1\times\text{C}}$
$=-\log\sqrt{1.80\times10^{-5}\times10^{-2}}$
$=-\log\sqrt{18\times10^{-8}}$
$=-\log4.24=0.62$
$\text{pH}=-\log4.24-\log10^{-4}$
$=0.62+4.00=3.38\simeq3.4$
View full question & answer→MCQ 181 Mark
At a temperature under high pressure $\mathrm{K}_{\mathrm{w}}\left(\mathrm{H}_2 \mathrm{O}\right)=10^{10}$, a solution of $\ce{pH}\ 5.4$ is said to be :
View full question & answer→MCQ 191 Mark
Solubility of calcium phosphate $($molecular mass $,M)$ in water is $\text{Wg}$ per $100mL$ at $25^\circ C$. Its solubility product at $25^\circ C$ will be approximately.
- A
$10^7\Big(\frac{\text{W}}{\text{M}}\Big)^3$
- ✓
$10^7\Big(\frac{\text{W}}{\text{M}}\Big)^5$
- C
$10^5\Big(\frac{\text{W}}{\text{M}}\Big)^5$
- D
$10^3\Big(\frac{\text{W}}{\text{M}}\Big)^5$
AnswerCorrect option: B. $10^7\Big(\frac{\text{W}}{\text{M}}\Big)^5$
$\text{S}=\frac{10\text{W}}{\text{M}}\text{moL/ L}$
$\text{K}_{\text{sp}}$ of $\text{Ca}_3(\text{PO}_4)_2=108\text{S}^5$
$=108\Big(\frac{10\text{W}}{\text{M}}\Big)^5$
$=10^7\Big(\frac{\text{W}}{\text{M}}\Big)^5\ ($approximately.$)$
View full question & answer→MCQ 201 Mark
At $363K,$ pure water has$[\mathrm{H}_2 \mathrm{O}^{+}]=10^{-6} \mathrm{M}$. The value of $\mathrm{K}_{\mathrm{w}}$ at this temperature will be :
- A
$10^{-6} $
- ✓
$10^{-12} $
- C
$ 10^{-13} $
- D
$ 10^{-14}$
AnswerCorrect option: B. $10^{-12} $
$[\text{H}_3\text{O}^+]=[\text{OH}^-]=10^{-6}\text{M}$
$\text{K}_{\text{w}}=[\text{H}_3\text{O}^+][\text{OH}^-]$
$=10^{-6}\times10^{-6}=10^{-12}\text{M}.$
View full question & answer→MCQ 211 Mark
A 0.2 molar solution of formic acid is 3.2% ionised. Its ionisation constant is:
- A
$9.6 \times 10^{-3} $
- ✓
$ 2.1 \times 10^{-4} $
- C
$ 1.25 \times 10^{-6} $
- D
$ 4.8 \times 10^{-5} $
AnswerCorrect option: B. $ 2.1 \times 10^{-4} $
$ 2.1 \times 10^{-4} $
Explanation:
$\text{K}=\text{C}\alpha^2=0.2\times(0.032)^2$
$=2.1\times10^{-4}$
View full question & answer→MCQ 221 Mark
For the reaction, $\text{CO(g)}+\text{ClO}_2\text{(g)}\rightleftharpoons\text{COCl}_2(\text{g}),$ the $\frac{\text{K}_{\text{p}}}{\text{K}_\text{C}}$ is equal to :
- A
$\sqrt{\text{RT}}$
- B
$\text{RT}$
- ✓
$\frac{1}{\text{R.T}}$
- D
$1.0$
AnswerCorrect option: C. $\frac{1}{\text{R.T}}$
$\text{K}_{\text{p}}=\text{K}_{\text{C}}(\text{RT})^{\Delta\text{n}}$
$\Rightarrow\text{K}_{\text{p}}=\text{K}_{\text{C}}(\text{RT})^{1-2}$
$\Rightarrow\frac{\text{K}_{\text{p}}}{\text{K}_{\text{C}}}=\frac{1}{\text{RT}}$
View full question & answer→MCQ 231 Mark
$100\ mL$ of a solution contains $0.1M\ \ce{NH_4OH}$ and $0.1M\ \ce{NH_4Cl}.$ The $\ce{pH}$ of the solution will not change on adding :
- A
$20\ mL$ of $0.1M\ \ce{NH_4OH}$ solution.
- B
$20\ mL$ of $0.1M\ \ce{NH_4Cl}$ solution.
- C
$10\ mL$ of $0.1M\ \ce{NaOH}$ solution.
- ✓
$10\ mL$ of distilled water.
AnswerCorrect option: D. $10\ mL$ of distilled water.
$\ce{pH}$ of buffer does not change on adding distilled water as there are no ions present in it.
View full question & answer→MCQ 241 Mark
Le Chatelier's principle is applicable to :
- A
Only homogeneous chemical reversible reaction.
- B
Only heterogeneous chemical reversible reaction.
- C
Only physical equilibria.
- ✓
All systems chemical or physical, in equilibrium.
AnswerCorrect option: D. All systems chemical or physical, in equilibrium.
Le Chatelier’s principle states that When some variables like temperature, pressure, concentration, etc are changed in a system at equilibrium then the system will go in that direction where it can get released from the change.
Le chatelier's principle is only applicable for the compounds at equilibrium.
So, Le Chatelier’s principle is applicable to all systems chemical or physical at equilibrium.
View full question & answer→MCQ 251 Mark
Which one of the following informations can be obtained on the basis of Le $-$ Chatelier principle?
- A
Dissociation constant of a weak acid.
- B
Entropy change in a reaction.
- C
Equilibrium constant of a chemical reaction.
- ✓
View full question & answer→MCQ 261 Mark
The addition of $\ce{HCl}$ will not suppress the ionisation of :
View full question & answer→MCQ 271 Mark
Which of the following is not affected by change in pressure?
- ✓
$ \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NO}(\mathrm{g}) $
- B
$ \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NH}_3(\mathrm{~g}) $
- C
$ \mathrm{PCl}_5(\mathrm{~g}) \Leftrightarrow \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2$
- D
$ 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{SO}_3(\mathrm{~g})$
AnswerCorrect option: A. $ \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NO}(\mathrm{g}) $
For the reaction, N$ \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NO}(\mathrm{g}) $, the value of $\triangle n$ is $2 − (1 + 1) = 0.$ When there is no change in the number of moles of reactants and products, there is no effect on the equilibrium when pressure is changed.
In all other option, the number of mole of reactant and products are different, Hence, if pressure is changed the equilibrium will be affected.
View full question & answer→MCQ 281 Mark
It accepts a proton. It is called as :
AnswerA Bronsted acid can accept a proton $\ce{(H^+)}.$ For example a basic salt, such as $\mathrm{Na}^{\oplus} \mathrm{H}^{\ominus}$, can take up a $\mathrm{H}^{+}$from $\mathrm{H}_2 \mathrm{O}$ to form $\mathrm{OH}^{\ominus}$ ions.
$\mathrm{F}^{\ominus}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HF}(\mathrm{aq})+\mathrm{OH}^{\oplus}(\mathrm{aq})$
View full question & answer→MCQ 291 Mark
$1M\ \ce{ NaCl}$ and $1M\ \ce{ HCl }$ are present in an aqueous solution . The solution is :
- ✓
Not a buffer solution with $\ce{pH } < 7$
- B
Not a buffer solution with $\ce{pH } > 7$
- C
A buffer solution with $\ce{pH } < 7$
- D
A buffer solution with $\ce{pH } > 7$
AnswerCorrect option: A. Not a buffer solution with $\ce{pH } < 7$
Buffer can accept and donate protons at the same time and $\ce{HC_1}$ is an acid.
So, it has $\ce{pH } < 7.$
So, this is not a buffer and the solution will be acidic.
View full question & answer→MCQ 301 Mark
Which buffer solution comprising of the following has its $\ce{pH }$ value greater than $7\ ?$
- A
$ \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa} $
- B
$ \mathrm{HCOOH}+\mathrm{HCOOK}$
- C
$ \mathrm{CH}_3 \mathrm{COONH}4 $
- ✓
$ \mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl} $
AnswerCorrect option: D. $ \mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl} $
View full question & answer→MCQ 311 Mark
Production of ammonia according to the reaction,
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})2\text{NH}_3(\text{g}); \Delta\text{H}=-92.38\text{kJ mol}^{-1}$ is an exothermic process. At low temperature, the reaction shifts in :
- ✓
- B
- C
Either forward or backward direction.
- D
View full question & answer→MCQ 321 Mark
The equibrium constant $\ce{K_c}$ for the reaction : $\text{H}_2(\text{g})+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI}(\text{g})\text{ at }700\text{K}\text{ is }49$ The $'K\ '$ for reaction $\text{HI(g)}\rightleftharpoons\frac{1}{2}\text{H}_2(\text{g})+\ \frac{1}{2}\text{I}_2(\text{g}).$
- A
$49$
- B
$0.02$
- ✓
$\frac{1}{7}$
- D
$1.43$
AnswerCorrect option: C. $\frac{1}{7}$
$\text{K}=\frac{[\text{KI}^2]}{[\text{H}_2][\text{I}_2]}=49$
$\Rightarrow\text{K}'=\frac{[\text{H}_2]^{\frac{1}{2}}[\text{I}^2]^{\frac{1}{2}}}{[\text{HI}]}$
$=\frac{1}{\sqrt{\text{K}}}=\frac{1}{\sqrt{49}}=\frac{1}{7}$
View full question & answer→MCQ 331 Mark
The dissociation constant of water is represented by $\text{K}=\frac{[\text{H}_3\text{O}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$$\text{or }=[\text{H}^+][\text{OH}^-]\text{K}_{\text{w}}$ is called:
- A
- ✓
- C
Ionisation constant of water.
- D
Ionisation constant of acid and base.
AnswerThe dissociation constant is represented by,
$\text{K}=\frac{[\text{H}_3\text{O}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$
The concentration of water is omitted from the denominator as water is a pure liquid and its concentration remains constant. $\ce{[H_2O)}$ is incorporated within the equilibrium constant to give a new constant, $K_w$, which is called the ionic product of water,
$\text{K}_{\text{w}}=[\text{H}^+][\text{OH}^-]$
View full question & answer→MCQ 341 Mark
The chemical equilibrium of reversible reaction is not influenced by:
- A
- ✓
- C
Concentration of the reactants.
- D
AnswerChemical equilibrium of reversible reaction is not influenced by the catalyst. Catalyst decreases the activation barrier for a reaction so the reaction proceeds fast. In the presence of the catalyst, the equilibrium reaches faster but it doesn't affect the thermodynamic properties.
View full question & answer→MCQ 351 Mark
At $55^{\circ}C$, autoprotolysis constant of water is $4 × 10^{-14}$. If a given sample of water has a $pH$ of $6.9$, then it is:
View full question & answer→MCQ 361 Mark
Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the:
- ✓
$pH$ scale.
- B
$\ce{pOH}$ scale.
- C
- D
Both $(a)$ and $(b).$
AnswerCorrect option: A. $pH$ scale.
View full question & answer→MCQ 371 Mark
$1\ mL$ of $\frac{\text{N}}{100}\text{HCl} $ is added to $1L$ of buffer having $pH = 5$. The $pH$ of the solution will be:
- A
Become $7$
- B
Become $3$
- C
Become $6$
- ✓
View full question & answer→MCQ 381 Mark
We know that the relationship between $K_c$ and $K_p$ is $\text{K}_\text{p}=\text{K}_\text{c}\text{(RT})^{\Delta\text{n}}$ What would be the value of $\Delta\text{n}$ for the reaction $\text{NH}_4\text{Cl(s)}\rightleftharpoons\text{NH}_3\text{(g)}+\text{HCl(g)}$
AnswerThe relationship between $K_p$ and $K_c$ is
$\text{K}_\text{p}=\text{K}_\text{c}\text{(RT)}^{\Delta\text{n}}$
Where $\Delta\text{n} = ($number of moles of gaseous products$) - ($number of moles of gaseous reactants$)$
For the reaction,
$\text{NH}_4\text{C1(s)}\leftrightharpoons\text{NH}_3\text{(g)}+\text{HCl(g)}$
$\Delta\text{n}=2-0=2$
View full question & answer→MCQ 391 Mark
For the reaction, $\text{H}_2(\text{g})+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI}(\text{g}),$ 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. $K < 1$
View full question & answer→MCQ 401 Mark
Ammonia dissociates to give nitrogen and hydrogen. What happens if the pressure is increased on the system at equilibrium?
- A
- ✓
- C
Backward reaction is exothermic.
- D
Answer$2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})$
upon increase in pressure the equilibrium will shift to the side of lesser gas molecules i.e. left. and thereby a decrease in number of moles of gas and volume of the system.
View full question & answer→MCQ 411 Mark
$\mathrm{PCl}_5, \mathrm{PCl}_3$ and $\mathrm{Cl}_2$ are at equilibrium at $500\ K$ in a closed container and their concentrations are $0.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}, 1.2 \times 10^{-3}$ $\mathrm{mol} \mathrm{L}^{-1}$ and $1.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$ respectively.
The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction $\text{PCl}_5\text{g}\rightleftharpoons\text{PCl}_3\text{(g)}+\text{Cl}_2\text{g}$ will be.
AnswerCorrect option: B. $1.8 \times 10^{-3}$
$\text{PCl}_5\text{g}\rightleftharpoons\text{PCl}_3\text{(g)}+\text{Cl}_2\text{g}$
$\text{K}_\text{c}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}=\frac{1.2\times10^{-3}\times1.2\times10^{-3}}{0.8\times10^{-3}}$
$[=1.8\times10^{-3}]\text{moL L}^{-1}$
View full question & answer→MCQ 421 Mark
In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?
- A
$\text{H}_2\text{(g)}+\text{I}_2\text{(g)}\rightleftharpoons\text{2HI(g)}$
- B
$\text{PCl}_5\text{(g)}\rightleftharpoons\text{PCl}_3\text{(g)}+\text{Cl}_2\text{(g)}$
- C
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)}$
- ✓
The equilibrium will remain unaffected in all the three cases.
AnswerCorrect option: D. The equilibrium will remain unaffected in all the three cases.
At constant volume equilibrium remains unaffected with the addition of inert gas.
View full question & answer→MCQ 431 Mark
Buffer Solution is prepared by mixing $.........$
- ✓
Weak acid and its salt of strong base.
- B
Strong acid $+$ its salt of strong base.
- C
Weak acid $+$ its salt of weak base.
- D
Strong base $+$ its salts of strong acid.
AnswerCorrect option: A. Weak acid and its salt of strong base.
A solution that resists change in $pH$ value upon addition of a small amount of strong acid or base $($less than $1 \%)$ or when the solution is diluted is called buffer solution.
An acidic buffer solution consists of a solution of a weak acid and its salt with a strong base. while basic buffer solution consists of a mixture of a weak base and its salt with strong acid.
View full question & answer→MCQ 441 Mark
The solubility product of $\mathrm{CaSO}_4$ is $6.4 \times 10^{-5}$. The solubility of salt in mol $L^{-1}$ is:
- A
$8.10^{-16}$
- ✓
$8.10^{-2}$
- C
$8.10^{-3}$
- D
$1.6^{-3}$
AnswerCorrect option: B. $8.10^{-2}$
$\text{CaSO}_4(\text{s})\rightleftharpoons\text{Ca}^{2+}(\text{aq})+\text{SO}^{2-}_4$
$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]$
$\Rightarrow6.4\times10^{-5}\times=\text{s}\times\text{s}$
$\Rightarrow\text{s}^2=\sqrt{64\times10^{-6}}$
$\Rightarrow\text{s}=8\times10^{-3}\text{mol L}^{-1}$
View full question & answer→MCQ 451 Mark
$100\ ml$ of a buffer of $1 \mathrm{MNH}_3(\mathrm{aq})$ and $1 \mathrm{MNH}_4^{+}(\mathrm{aq})$ are placed in two half cells separately. A current of $1.5\ amp$ is passed through both half cells for $20$ min, if electrolysis of only water take place. The $pH$ of :
- ✓
Anode half cell will decrease.
- B
Cathode half cell will decrease.
- C
Both half cell will decrease.
- D
Both half cell remain same.
AnswerCorrect option: A. Anode half cell will decrease.
As only the electrolysis of water take place so the conc. of $H^+$ ion in Anode compartment increases.
View full question & answer→MCQ 461 Mark
- A
$ \mathrm{NH}_4 \mathrm{C} 1+\mathrm{HC} 1 $
- B
$ \mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{CO}_3$
- ✓
$40\ ml$ of $\ce{0.1 M NaCN + 20ml}$ of $\ce{0.1 MHCN}$
- D
$\ce{NaC1 + NaOH}$
AnswerCorrect option: C. $40\ ml$ of $\ce{0.1 M NaCN + 20ml}$ of $\ce{0.1 MHCN}$
Buffer solution is a mixture of weak acid and its salt of weak acid.
$\ce{NaCN} \rightarrow$ salt of strong base and weak acid
$\ce{HCN} \rightarrow$ weak acid
$\therefore$ They can act as buffer.
While no other options follow definition of buffer solution.
View full question & answer→MCQ 471 Mark
The product of molar concentrations of hydrogen ions and hydroxide ions in a $0.01\ M$ aqueous solution of sodium chloride is known as:
- A
Hydrolysis constant of salt.
- B
Dissociation constant of acid.
- C
Dissociation constant of base.
- ✓
AnswerSodium chloride is a salt of strong base $\text{NaOH}$ and strong acid $\text{HCl}$. In its aqueous solution, following equilibrium is observed.
$\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$
View full question & answer→MCQ 481 Mark
In the gaseous equilibrium $A + 2B ⇌ C +$ Heat, the forward reaction is favoured:
- A
Low $P$, High $T$
- B
Low $P$, Low $T$
- ✓
High $P$, Low $T$
- D
High $P$, High $T$
AnswerCorrect option: C. High $P$, Low $T$
$A + 2B ⇌ C +$ Heat
The equation shows, that it is exothermic reaction.
Since the heat is released in the reaction, so the reaction is favoured in forward direction at low temperature.
$\triangle n = 1 − (2 + 1) = −2$
View full question & answer→MCQ 491 Mark
An equimolar solution of $\ce{NaNO_2}$ and $\ce{HNO_2}$ can act as a:
Answer$\ce{HNO_2}$ is a weak acid.
$\ce{NaNO_2}$ is a salt of weak acid $\ce{HNO_2}$ with strong base $\ce{NaOH}.$
View full question & answer→MCQ 501 Mark
At $500K$, equilibrium constant, $K_c$ , for the following reaction is $5.\ \frac{1}{2}\text{H}_2\text{(g)}+\frac{1}{2}\text{I}_2\text{(g)}\rightleftharpoons\text{HI(g)}$ What would be the equilibrium constant $K_c$ for the reaction $2\text{HI}\text{(g)}\rightleftharpoons\text{H}_2\text{(g)}+\text{I}_2\text{(g)}$
AnswerCorrect option: A. $0.04$
For the reaction,
$\frac{1}{2}\text{H}_2\text{(g)}+\frac{1}{2}\text{I}_2\text{(g)}\rightleftharpoons\text{HI(g)}\ ;\text{K}_\text{c}=5$
For $\text{H}_2\text{(g)}+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI}\text{(g)}\ ;\text{K}_\text{c}=(5)^2=25$
For $2\text{HI}\text{(g)}\rightleftharpoons\text{H}_2\text{(g)}+\text{I}_2\text{(g)}\ ;\text{K}_\text{c}=\frac{1}{25}=0.04$
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