If $$1M$$ $$C{H_3}COONa$$ is added to $$1M$$ $$C{H_3}COOH$$, then:

$$pH$$ of the solution increases

MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 14

$90^{ \circ }C$ , pure water has

$[{ H }_{ 3 }O^{ + }]={ 10 }^{ -6.7 }\ mol/lit$ . What is the value of

${ K }_{ W }$ at

${ 90 }^{ \circ }C :-$

0%
${ 10 }^{ -6 }$
0%
${ 10 }^{ -12 }$
0%
${ 10 }^{ -14 }$
0%
${ 10 }^{ -13.4 }$

Which of the following will not function as buffer solution?

0%
${\text{Na}}{{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ + N}}{{\text{a}}_{\text{2}}}{\text{HP}}{{\text{O}}_{\text{4}}}\left( {{\text{1:1}}\,{\text{Molar}}\,{\text{Ratio}}} \right)$
0%
${\text{N}}{{\text{H}}_4}{\text{Cl + NaOH}}\left( {{\text{2:1}}\,{\text{Molar}}\,{\text{Ratio}}} \right)$
0%
${\text{HCOOH + NaOH}}\left( {{\text{1:1}}\,{\text{Molar}}\,{\text{Ratio}}} \right)$
0%
${\text{HCOOH + HCOONa}}\left( {{\text{1:1}}\,{\text{Molar}}\,{\text{Ratio}}} \right)$

In which of the following solution of electrolyte, ionic equilibrium is set up?

0%
$H_{2}SO_{4}$
0%
$HCl$
0%
$NaOH$
0%
$HCN$

Which of the following is a buffer solution?

0%
500 mL of 0.1 N CH $_3$ COOH +500 mL of 0.4 N NaOH
0%
500 mL of 0.1 N CH $_3$ COOH +500 mL of 0.1 N HCl
0%
500 mL of 0.1 N CH $_3$ COOH +500 mL of 0.2 N NaOH
0%
500 mL of 0.1 N CH $_3$ COOH +500 mL of 0.3N NaOH

On adding NaCl solution to the saturated solution of

$PbCl_2$ then the concentration of :

0%
$Pb^{2+}$ will increase
0%
$Cl^-$ will decrease
0%
$Pb^{2+}$ will decrease
0%
Ions will remain unaltered

The ionic product of water at

$25^oC$ is

$10^{-14}$ The ionic product at

$90^oC$ will be

0%
$1 \times 10^{-20}$
0%
$1 \times 10^{-12}$
0%
$1 \times 10^{-14}$
0%
$1 \times 10^{-16}$

In a buffer solution the ratio of concentration of

${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\,\,{\text{and}}\,{\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}\,\,{\text{is}}\,\,{\text{1:1}}$ when it changes in 2:1 what will be the value of pH of buffer:-

0%
Increase
0%
Decrease
0%
No effects
0%
N. O. T.

A hydrogen electrode placed in a buffer solution of NaCl and HCN in the ratio of X:Y and Y:X has an oxidation electrode potential values

${E_1}\,\,and\,\,{E_2}$ volts respectively at

${25^0}C$ . The pKa value of HCN is

0%
$\frac{{({E_1} + {E_2})}}{{0.118}}$
0%
$- \frac{{({E_1} + {E_2})}}{{0.118}}$
0%
$\frac{{({E_1} - {E_2})}}{{0.118}}$
0%
$\frac{{({E_2} - {E_1})}}{{0.118}}$

If the degree of ionization of water is 1.8 x

$10^{-9}$ at a given temperature. Its ionic product will be

0%
1.8 x $10^{-16}$
0%
1 x $10^{-14}$
0%
1 x $10^{-16}$
0%
1.67 x $10^{-14}$

$2{H_2}O\leftrightharpoons {H_3}{O^ + } + O{H^ - },{K_w} = 1\times {10^{ - 14}}$ at

${25^ \circ }C$ hence

$K_a$ is :

0%
$1\times 10^{-14}$
0%
$18\times 10^{-17}$
0%
$5.55\times 10^{-3}$
0%
$1.00\times 10^{-7}$

The dissociation constant of a weak acid is

$1.0 \times { 10 }^{ -5 }$ , the equilibrium constant for its reaction with strong base is

0%
$1.0 \times { 10 }^{ -5 }$
0%
$1.0 \times { 10 }^{ 9 }$
0%
$1.0 \times { 10 }^{ 7 }$
0%
$1.0 \times { 10 }^{ 14 }$

One litre of water contains

$10 ^ { - 7 }$ mole of

$H ^ { + }$ ions. Degree of ionisation of water is:

0%
$1.8 \times 10 ^ { - 7 } \%$
0%
$0.8 \times 10 ^ { - 9 } \%$
0%
$3.6 \times 10 ^ { - 9 } \%$
0%
$3.6 \times 10 ^ { - 7 } \%$

Explanation

$1$ litre of water contains

$1000/18$ mole.

So degree of ionization

$=\dfrac {10^{-7}\times 18}{1000}=1.8\times 10^{-7}\%$

Option A is correct.

Aqueous solution of NaCl is neutral because

0%
$Na^{ + }$ undergoes hydrolysis
0%
Cl undergoes hydrolysis
0%
Both $Na^{ + }$ and Cl undergo hydrolusis
0%
Does not undergo hydrolysis

Which of the following may be regarded as buffer solution?

0%
50 ml of 0.1 M acetic acid + 50 ml of 0.05 M $NaOH$
0%
50 ml of 0.05 ammonia + 50 ml of 0.1 M $HCl$
0%
25 ml of 0.1 M acetic acid + 50 ml of 0.1 M $NaOH$
0%
50 ml of 0.1 ammonia + 50 ml of 0.05 M $HCl$

The total number of different kind of buffer obtained during the titration of

${ H }_{ 3 }{ PO }_{ 4 }$ with NaOH are:-

0%
3
0%
1
0%
2
0%
0

Select the correct acid-base equilibrium.

0%
$\mathrm { HNO } _ { 3 } + \mathrm { HF } \rightleftharpoons \mathrm { H } _ { 2 } \mathrm { NO } _ { 3 } ^ { + } + \mathrm { F } ^ { - }$
0%
$\mathrm { HNO } _ { 3 } + \mathrm { HF } \rightleftharpoons \mathrm { H } _ { 2 } \mathrm { F } ^ { + } + \mathrm { NO } _ { 3 } ^ { - }$
0%
$\mathrm { CH } _ { 3 } \mathrm { COOH } + \mathrm { NH } _ { 3 } \rightleftharpoons \mathrm { CH } _ { 3 } \mathrm { COOH } _ { 2 } ^ { + } + \mathrm { NH } _ { 2 } ^ { + }$
0%
$\mathrm { HF } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { H } _ { 2 } \mathrm { F } ^ { + } + \mathrm { OH } ^ { - }$

$HCOOH$ and

$CH_3COOH$ solutions have equal

$pH$ . If

$\cfrac{K_1}{K_2}$ (ratio of ionisation constants of acids) is 4, their molar concentration ratio will be:

0%
2
0%
0.5
0%
4
0%
0.25

$90 ^ { \circ } C ,$ pure water has

$\left[ \mathrm { H } _ { 3 } \mathrm { O } ^ { + } \right] = 10 ^ { - 6 } \mathrm { mol }$ litre

$^ { - 1 } .$ The value of

$K _ { w }$ at

$90 ^ { \circ } \mathrm { C }$ is:

0%
$10 ^ { - 6 }$
0%
$10 ^ { - 12 }$
0%
$10 ^ { - 14 }$
0%
$10 ^ { - 8 }$

Hydrogen ion concentration of an aqueous solution is

$1 \times 10^{-4}M$ . The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol

$dm^{-3}$ is __________.

0%
$1 \times 10^{-8}$
0%
$1 \times 10^{-6}$
0%
$2 \times 10^{-10}$
0%
$0.5 \times 10^{-10}$

Explanation

$\text{Concentration of aqueous solution=} 10^{−4}M$

$\text{If we add equal volume, let say V of water. The final volume of solution will be 2V. The new concentration is}$

$⇒M_1V_1=M_2V_2$

$⇒10^{−4}M×V=M_2×2V$

$⇒M_2=5×10−5M ⟶ \text{New concentration of }[H^+] .$

$pH=−log[H^+]=−log[5×10^{−5}]$

$⇒pH=4.3$

$⇒pOH=14−pH=9.69$

$⇒−log[OH^-]=9.69$

$⇒[OH^-]=10^{−9.69}=2×10^{−10}$

$\text{Thus the concentration of }[OH^-] \ is\ 2×10^{−10}M .$

$1M$

$C{H_3}COONa$ is added to

$1M$

$C{H_3}COOH$ , then:

0%
$pH$ of the solution increases
0%
$pH$ decreases
0%
$pH$ does not change
0%
cannot be predicted

The dissociation constants for acetic acid and HCN at

$25^{0}$ C are

$1 . 5 \times 10^{-5}$ and

$4 . 5 \times 10 ^{-10}$ , respectively. The equilibrium constant for the equilibrium

$CN^{-} + CH_{3}COOH \leftrightharpoons HCN + CH_{3}COO^{-}$ would be :

0%
$3\times10^{4}$
0%
$3\times10^{5}$
0%
$3\times10^{-5}$
0%
$3\times10^{-4}$

${ 25 }^{ \circ }C$ the dissociation constant of a base , BOH, is

$1.0\times { 10 }^{ -12 }$ . The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be

0%
$1.0\times { 10 }^{ -6 }{ molL }^{ -1 }$
0%
$1.0\times { 10 }^{ -7 }{ molL }^{ -1 }$
0%
$2.0\times { 10 }^{ -6 }{ molL }^{ -1 }$
0%
$1.0\times { 10 }^{ -5 }{ molL }^{ -1 }$

A solution is saturated with respect to

$AgSCN$ as well as

$AgBr$ . The concentration of

$Ag^{+}$ in the solution would be:

(

$K_{sp}$ for

$AgSCN = 1.2 \times 10^{-12}$ and for

$AgBr = 5 \times 10^{-13}$ )

0%
$1.3 \times 10^{-5}$
0%
$1.3 \times 10^{-12}$
0%
$1.3 \times 10^{-6}$
0%
$1.3 \times 10^{-8}$

Which of the following salt evolve(s)

${ N }_{ 2 }$ gas on heating ?

0%
$\left( { NH }_{ 4 } \right) _{ { 2 } }{ CrO }_{ 4 }$
0%
${ NH }_{ 4 }{ NO }_{ 2 }$
0%
${ NH }_{ 4 }{ NO }_{ 3 }$
0%
$\left( { NH }_{ 4 } \right) _{ { 2 } }{ CO }_{ 3 }$

Which of following salt solution will impart greenish colour ?

0%
$K_2Cr_2O_7$
0%
$ZnCl_2$
0%
$K_2MnO_4$
0%
$K_2CrO_4$

10 ml of 0.2 M

$AgNO_3$ is mixed with 10 ml of 0.1 M

$NaCl$ . What is the concenctration of

$Cl^-$ in the resulting solution

$[ K_{sp} (AgCl) = 10^{-10} M^2]$

0%
$10^{-5} M$
0%
$10^{-9} M$
0%
$2 \times 10^{-9} M$
0%
$10^{-8} M$

For preparing a buffer solution of pH = 7.0, which buffer system you will choose?

0%
$H_3PO_4,H_2PO_4^-$
0%
$H_2PO_4^-, HPO_4^{2-}$
0%
$HPO^{2-}_4, PO^{3-}_4$
0%
$H_3PO_4, PO_4^{3-}$

The percentage of pyridine

$\left( { C }_{ 5 }{ H }_{ 5 }N \right)$ that forms pyridinium ion

$\left( { C }_{ 5 }{ H }_{ 5 }{ N }^{ + }H \right)$ in a 0.10 m aqueous pyridine solution

$\left( K_{ b } \ for\ { C }_{ 5 }{ H }_{ 5 }N=1.7\times { 10 }^{ -9 } \right)$ is:

0%
0.0060%
0%
0.013%
0%
0.77%
0%
1.6%

A weak acid react with strong base, ionisation constant of weak acid is

${ 10 }^{ -4 }$ . Find out equilibrium constant for this reaction.

0%
${ 10 }^{ -10 }$
0%
${ 10 }^{ 10 }$
0%
${ 10 }^{ -9}$
0%
${ 10 }^{ 9}$

Which statement is/are correct-

0%
If $0.1\ M$ $CH_3COOH$ aqueous solution $( K_a = 1.8 \times 10^{-5} )$ is diluted at $2560^o C$ , then $[H^+]$ will increase
0%
On heating a sample of pure water $[H^+]$ and $[OH^-]$ increases but the sample still remains neutral
0%
$10^{-4}\ M$ $HCl$ solution is less acidic than $0.1\ M$ $HCN$ $( K_a HCN = 10^{-5} )$
0%
$10^{-8}\ M$ $HCl$ aqueous solution will have pH exactly equal to 8

The self ionisation constant of

$\mathrm { NH } _ { 3 }$ at

$50 ^ { \circ } \mathrm { C }$ is given by

$\mathrm { K } _ { \mathrm { NH } _ { 3 } } = \left[ \mathrm { NH } _ { 4 } ^ { + } \right] \left[ \mathrm { NH } _ { 2 } ^ { - } \right] = 10 ^ { - 30 }$ . How many

$N H _ { 2 } ^ { - }$ ions are present per

$\mathrm { cm } ^ { 3 }$ of pure liquid

$\mathrm { NH } _ { 3 }$ ?

0%
$6 \times 10 ^ { 6 }$
0%
$6 \times 10 ^ { 5 }$
0%
$6 \times 10 ^ { - 5 }$
0%
$6 \times 10 ^ { - 6 }$

The dissociation constant of an acid HA is

$1 \times 10 ^ { - 5 }$ . The pH of 0.1 molar solution of the acid will be approximately:

0%
3
0%
5
0%
1
0%
6

The silicate anion in the mineral kinetic is a chain of three

$SiO_4^{4^-}$ tetrahedral that shares conners with adjacent tetrahedral. The mineral also contains

$Ca$ icon

$Cu$ ions and water molecules in a 1:1:1 ratio mineral represented as:

0%
$CaCu_2Si_{ 3 }O_{ 10 }.H_{ 2 }O$
0%
$CaCuSi_{ 3 }O_{ 10 }.H_{ 2 }O$
0%
$Ca_{ 2 }Cu_{ 2 }Si_{ 3 }O_{ 10 }.2H_{ 2 }O$
0%
None of these

Which of the following is ionic?

0%
$HCl$
0%
$CHCl_3$
0%
$IF_5$
0%
$KI$

For which salt the

$\mathrm { pH }$ of its solution does not change with dilution?

0%
$\mathrm { NH } _ { 4 } \mathrm { Cl }$
0%
$\mathrm { CH } _ { 3 } \mathrm { COONH } _ { 4 }$
0%
$\mathrm { CH } _ { 3 } \mathrm { COONa }$
0%
None of these

Consider the following equilibrium.
Liquid

$\rightleftharpoons$ Vapour
Which of the following relation is correct?

0%
$\dfrac{d ln P}{dT}=\dfrac{\Delta Hv}{RT^2}$
0%
$\dfrac{d ln G}{dT^2}=\dfrac{\Delta Hv}{RT^2}$
0%
$\dfrac{d ln P}{dT}=\dfrac{-\Delta Hv}{RT}$
0%
$\dfrac{d ln P}{dT^2}=\dfrac{-\Delta Hv}{T^2}$

The

$pH$ of an aqueous solution of sodium chloride at

$60^{\circ}C$ is

0%
$7.0$
0%
$> 7.0$
0%
$<7.0$
0%
$0$

Explanation

We know that at

$25^{\circ}C$ dissociation constant

$K_w$ of water is,

$H_2O\rightleftharpoons H^++OH^-$

$[H^+][OH^-]=10^{-14}=K_w$

So,

$[H^+]=[OH^-]=10^{-7}M$

Thus

$pH=7$ for

$H_2O$ at

$25^{\circ}C$

Adding

$NaCl$ does not affect the

$pH$ whatever be the temperature because sodium chloride is salt of strong acid and strong base. So, it is

$H^+$ ion of water which is affecting the

$pH$ .

Dissociation of water is an endothermic process. With an increase in temperature dissociation constant of water increases. So the dissociation constant is more i.e.,

$K_w>10^{-14}$ .

Therefore,

$[H^+]>10^{-7}$ and

$pH$ is less than 7.

Hence the correct option is (C).

The EMF of cell:

$H_{2}(g)|$ Buffer

$\parallel$ Normal calmelelectrod, is 0.70 V at

$25^{\circ}C,$ when barometric pressure is 760 mm. What is the pH of the buffer solution?

$E_{Calomel}^{0} = 0.28 V. [2.303 RTIF = 0.06]$

0%
3.5
0%
7.0
0%
tending to zero
0%
tending to 14.0

The dissociation constant of formic acid is

$0.00024$ . The hydrogen ion concentration in

$0.002 M \ HCOOH$ solution is nearly

0%
$6.93 \times 10^{-4}M$
0%
$4.8 \times 10^{-7}M$
0%
$5.8 \times 10^{-4}M$
0%
$1.4 \times 10^{-4}M$

The degree of dissociation of pure water at

$25^{\circ}C$ is found to be

$1.8 \times 10^{-9}$ . The dissociation constant,

$K_{d}$ of water, at

$25^{\circ}C$ is

0%
$10^{-14}$
0%
$1.8 \times 10^{-16}$
0%
$5.56 \times 10^{-13}$
0%
$1.8 \times 10^{-14}$

Explanation

Given:

Degree of dissociation of pure water,

$\alpha =1.8\times 10^{-9}$

We know that,

Density of pure water =

$1\space g\space cm^{-3}=1000\space g\space L^{-1}$

Molar mass of water =

$18\space g\space mol^{-1}$

$Concentration=\frac{number\space of\space moles}{Volume}=\frac{1000\space g\space L^{-1}}{18\space g\space mol^{-1}}=55.5\space M$

Dissociation of water,

$H_2O\rightleftharpoons H^++OH^-$

Dissociation constant,

$K_d=\frac{[H^+][OH^-]}{[H_2O]}=\frac{C\alpha\times C\alpha}{C-C\alpha}$

$\alpha<<<1$ ,

$[H_2O]$ is not affected.

Therefore,

$K_d=c\alpha^2=55.5\times (1.8\times 10^{-9})^2=1.79\times 10^{-16}\approx 1.8\times 10^{-16}$

Hence, the correct option is (B).

Which one of the following salts gives aqueous solution which is neutral?

0%
$NaHC{ O }_{ 3 }$
0%
$NaHS{ O }_{ 4 }$
0%
$NaCl$
0%
$\\ Na{ H }_{ 4 }HC{ O }_{ 3 }$

When ammonia is added to pure water, the concentration of which of the following species already present, will decrease?

0%
$OH^-$
0%
$H_3O^+$
0%
$NH_4^+$
0%
None of these

n-caproic acid,

$C_{5}H_{11}COOH$ , found in coconut and palm oil is used in making artificial flavors, has solubility in water equal to

$11.6 g/L$ . The saturated solution has

$pH = 3.0$ . The

$K_{a }$ of acid is

0%
$10^{-6}$
0%
$10^{-5}$
0%
$2 \times 10^{-5}$
0%
$2 \times 10^{-6}$

What is

$[H^+]$ in mol/L of a solution that is 0.20 M

$CH_3CCONa$ and 0.10 M in

$CH_3COOH$ ?

$( K_a\ for\ CH_3COOH = 1.8 \times 10^{-5} )$

0%
$3.5 \times 10^{-4}$
0%
$1.1 \times 10^{-5}$
0%
$1.8 \times 10^{-5}$
0%
$9.0 \times 10^{-6}$

A solution has initially

$0.1 M -HCOOH$ and

$0.2 M- HCN$ .

$K_{a}$ of

$HCOOH = 2.56 \times 10^{-4}$ ,

$K_{a}$ of

$HCN = 9.6 \times 10^{-10}$ . The only incorrect statement for the solution is

$(log 2 = 0.3)$

0%
$[H]^{+} = 1.6 \times 10^{-3} M$
0%
$[HCOO]^{-} = 1.6 \times 10^{-3} M$
0%
$[CN]^{-} = 1.2 \times 10^{-7} M$
0%
$pOH = 2.8$

Calculate the formation constant for the reaction of a tripositive metal ion with thiocyanate ions to form the monocomplex if the total metal concentration in the solution is

$2 \times 10^{-3} M$ , the total

$SCN^{-}$ concentration is

$1.51 \times 10^{-3} M$ and the free

$SCN^{-}$ concentration is

$1.0 \times 10^{-5} M$ .

0%
$7.55 \times 10^{4}$
0%
$3 \times 10^{5}$
0%
$3.33 \times 10^{-6}$
0%
$1.5 \times 10^{5}$

Small amount of freshly precipitated magnesium hydroxides are stirred vigorously in a buffer solution containing

$0.25 M$ of

$NH_{4}Cl$ and

$0.05 M$ of

$NH_{4}OH$ .

$[Mg^{2+}]$ in the resulting solution is (

$K_{b}$ for

$NH_{4}OH = 2.0 \times 10^{-5}$ and

$K_{sp}$ of

$Mg(OH)_{2} = 8.0 \times 10^{-12}$ )

0%
$4 \times 10^{-6} M$
0%
$2 \times 10^{-6} M$
0%
$0.5 M$
0%
$2.0 M$

A volume of

$1.0 L$ of solution which was in equilibrium with solid mixture of

$AgCl$ and

$Ag_{2}CrO_{4}$ was found to contain

$1 \times 10^{-4}$ moles of

$Ag^{+}$ ions,

$1.0 \times 10^{-6}$ moles of

$CI^{-}$ ions and

$8.0 \times 10^{-4}$ moles of

$CrO_{4}^{2-}$ ions.

$Ag^{+}$ ions are added slowly to the above mixture (keeping the volume constant) till

$8.0 \times 10^{-7}$ moles of

$AgCl$ got precipitated. How many moles of

$Ag_{2}CrO_{4}$ were precipitated simultaneously?

0%
$7.68 \times 10^{-4}$
0%
$4.8 \times 10^{-4}$
0%
$8.0 \times 10^{-4}$
0%
$7.68 \times 10^{-5}$

Which one of the following is not an acid salt?

0%
$NaH_2PO_2$
0%
$NaH_2PO_3$
0%
$NaH_2PO_4$
0%
$Na_2HPO_4$

Explanation

$NaH_2PO_2$ is not an acidic salt. It is a basic salt. It is a salt of weak acid

$H_3PO_2$ and a strong base

$NaOH$ . Hence, the aqueous solution of

$NaH_2PO_2$ is alkaline in nature. In

$NaH_2PO_2,$ there is no acidic hydrogen.

Hence, the correct option is (A).

The solubility of

$AgCN$ in a buffer solution of

$pH - 3.0$ is (

$K_{sp}$ of

$AgCN = 1.2 \times 10^{-16}$ ;

$K_{a}$ of

$HCN = 4.8 \times 10^{-10}$ )

0%
$1.58 \times 10^{-5}M$
0%
$2.0 \times 10^{-5}M$
0%
$1.58 \times 10^{-4}M$
0%
$2.5 \times 10^{-9}M$

Explanation

Assume solubility of AgCN be a M in presence of a buffer of pH = 3

Given :-

$K_{sp}$ of AgCN =

$1.2 \times 10^{-16}$ and

$K_{a}$ for HCN =

$4.8 \times 10^{-10}$

$AgCN \rightleftharpoons Ag^{+} + CN^{-}$

$[Ag^{+}] = a; \quad [H^+] = 10^-{3}M;$

For weak HCN,

$K_{a} = \frac{[H^{+}][CN^{-}]}{[HCN]}$

$\Rightarrow 4.8 \times 10^{-10} = \dfrac{[10^{-3}][CN^{-}]}{a}$

$\therefore [CN^{-}] = 4.8 \times 10^{-7} \times a$

$For \ AgCN : K_{sp} = Ag^{+} \times CN^{-}$

$1.2 \times 10^{-16} = a \times 4.8 \times 10^{-7} \times a$

$a^{2} = \dfrac{1.2 \times 10^{-16}}{4.8 \times 10^{-7}}$

$a = 1.58 \times 10^{-5} M$

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 14 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 14 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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