MCQ Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 9

The electronic configuration of an element is

$1s^2, 2s^2, 2p^6, 3s^23p^3$ . What is the atomic number of the element which is just below the given element in the periodic table?

0%
$34$
0%
$49$
0%
$33$
0%
$31$

Explanation

The element with the given electronic configuration is P (Phosphorus) and the atomic number is 15 below this is atomic number 33 As (Arsenic).

The electronic configuration of the Arsenic is given as follow:

$[Ar]3d^{10}4s^24p^3$

$12\ g$ of a trivalent element combines with

$4\ g$ of oxygen. Its atomic mass is:

0%
$48$
0%
$72$
0%
$3$
0%
$36$

Explanation

Equivalent mass of metal

$= \dfrac{\text{Weight of metal}}{\text{Weight of oxygen}} \times 8$

$= \dfrac{12}{8} \times 8$

$= 12$

Therefore, atomic mass = Equivalent weight

$\times$ valency

$= 12 \times 3$

$= 36 gm$

The orbit having Bohr radius equal to

$1^{st}$ Bohr orbit of

$H-$ atom is:

0%
$n=2$ of $He^{+}$
0%
$n=2$ of $B^{+4}$
0%
$n=3$ of $Li^{+2}$
0%
$n=2$ of $Be^{+3}$

Explanation

$=\dfrac { { n }^{ 2 }{ h }^{ 2 } }{ 4{ \pi }^{ 2 }{ m }{ e }^{ 2 } } \times \dfrac { 1 }{ z }$

$0.529\times \dfrac { { n }^{ 2 } }{ z }$

where n=Orbit number

z=atomic number

n=1 of H atom

$=\dfrac{1}{1}=1$

For

$n=2$ of

${B}^{+4}$

$=\dfrac{4}{5}$

For

$n=2$ of

${He}^{+}$

$=\dfrac{4}{2}=2$

For

$n=3$ of

${Li}^{+2}$

$=\dfrac{9}{3}=3$

For

$n=2$ of

${Be}^{+2}$

$=\dfrac{4}{4}=1$

So radius of 1 st orbit of H atom is equal to

$n=2$ of

${Be}^{+3}$ atom

The number of orbitals in n = 3 are:

0%
1
0%
4
0%
9
0%
16

Explanation

There are nine orbitals in the n = 3 shell. There is one orbital in the 3s subshell and three orbitals in the 3p subshell. n = 3 shell, however, also includes 3d orbitals.

Also, we have to note that the number of orbitals is given by

$n^2$

The active mass of

$7.0\ g$ of nitrogen in a

$2.0\ L$ container would be:

0%
$0.25$
0%
$0.125$
0%
$0.5$
0%
$14.0$

Explanation

Moles of

$N_2=\dfrac{7}{28} = 0.25\ mol$

Active mass = conc. in

$mol/L$

$0.25 mol / 2 L$

= $0.125$

So, the correct option is $B$

The key feature Of Bohr's theory Of spectrum Of hydrogen atom is the quantization Of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy Of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.A diatomic molecule has moment Of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n=0 is not allowed) is

0%
$\cfrac{h^2}{(n^2)(8\pi^2I)}$
0%
$\cfrac{h^2}{(n)(8\pi^2I)}$
0%
$\cfrac{n h^2}{8\pi^2I}$
0%
$\cfrac{n^2 h^2}{(8\pi^2I)}$

Explanation

$\begin{array}{l}\text { Given - * A Diatomic molecule- } \\\text { let, the structure of the molecule be like- }\end{array}$

$\Rightarrow I=2 m r^{2}$

$\begin{array}{l}\text { we howe Rotational encrgy- } \\\qquad E=\frac{1}{2} I \omega^{2}=m \omega^{2} r^{2}-(1) \\\text { By Bohr quantization of angular momentum } \\\qquad 2 \text { mwr}^2=\frac{n h}{2 \pi} \quad-(2) \\\text { From } e q(1) \text { and (2) we get ,}\\\qquad E=\frac{n^{2} h^{2}}{16 \pi^{2} m r^{2}}=\frac{n^{2} h^{2}}{8\pi^{2} I}\end{array}$

1994219_1068496_ans_86b7fb048fa04ad6bc709fc0c20ec337.JPG

$7.8$ grams of metal in water and liberated

$2.24L$ of hydrogen at standard conditions. Calculate the atomic number of the metal.

0%
$23$
0%
$24$
0%
$7$
0%
$19$
0%
$35$

Which of the following statement is not correct?

0%
Atomic no.of A,B,C are $40,57,105$ respectively
0%
Group no of A, B,C are IV B, III B and V B respectively
0%
Period no of A, B,C are 4th , 5th and 6th respectively
0%
C is a radioactive element

Explanation

(Refer to Image)

Every element in the periodic table is arranged according to its atomic number.
So, according to the arrangement of an element in the periodic table.

Atomic number of

$A\longrightarrow 40, B\longrightarrow 57, C\longrightarrow 105$

1103033_1093820_ans_61c9e6e8170a4cac908e3751b92daa0d.png

Moseley's equation is represented as

$\sqrt { v } =a(Z-b)$ where a and b are constants If

$OA=1$ , then atomic number of the elements showing frequency of

$400Hz$ is?

0%
21
0%
11
0%
401
0%
19

Explanation

Given:

$OA=1\\V=400Hz\\a=1$

$\sqrt{V}=a(z-b),\\\sqrt{400}=1(z-1)\\20=z-1\\z=21$

Hence atomic number is

$21$ .

1171396_1096661_ans_cc624874703d4e8ab50af89d121b9b3a.png

Visible spectrum lies between:

0%
400nm to 760 nm
0%
350nm to 760 nm
0%
400nm to 660 nm
0%
300nm to 760 nm

What is the potential energy of an electron present in N-shell of the

${ B }e^{ 3+ } ion$ ?

0%
$-3.4 eV$
0%
$-6.8 eV$
0%
$-13.6 eV$
0%
$-27.2 eV$

Explanation

Energy of hydrogen like species

$= -13.6 \times \cfrac{{Z}^{2}}{{n}^{2}}$

Whereas,

$Z =$ Atomic number and

$n =$ Shell

$n$ shell represents

$n = 4$

$\therefore$ Energy of

${Be}^{+3}$ ion

$\left( E \right) = -13.6 \times \cfrac{{4}^{2}}{{4}^{2}} = -13.6 \; eV \; \left( \because {Z}_{Be} = 4 \right)$

$\therefore$ Potential energy

$\left( P.E. \right) = 2 \times E = 2 \times \left( -13.6 \right) = -27.2 \; eV$

Hence, the correct answer is

$D$

The orbit from which when electron will jump in other orbit, energy may be absorbed but not emitted out, will be?

0%
$1$ st orbit
0%
$2$ nd orbit
0%
$7$ th orbit
0%
infinite orbit

The second is the duration of

$9 192 631 770$ periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium

$133$ atom.

0%
True
0%
False

What atomic number of an element

$X$ would have to become so that the

$4th$ orbit around

$X$ would fit inside the

$1st$ Bohr orbit of

$H-$ atom?

0%
3
0%
4
0%
16
0%
25

Explanation

By definition of Bohr radius

$a_o$ ,

$r=n^2\left(\cfrac {a_o}{Z}\right), r=$ radical distance from the nucleus of a

$H$ -like atom

So, if we want

$r=a_o$ ,

$Z=n^2\left(\cfrac {a_o}{r}\right)=n^2\left(\cfrac {a_o}{a_o}\right)=n^2$

For the

$4^{th}$ Bohr orbit,

$n=4$ ,

$\Rightarrow Z= 4^2= 16$

What is the seperation energy (in eV) for

$Be^{3+}$ in the first excited state in eV?

0%
$13.6eV$
0%
$27.2eV$
0%
$40.8eV$
0%
$54.4eV$

Explanation

Separation Energy=

$13.6 \times \cfrac {Z^2}{n^2}$

For

$Be$ ,

$Z=4$

For

$1^{st}$ excited state

$n=2$

$\therefore 13.6 \times \cfrac {4^2}{2^2}= 54.4eV$

Suppose that A and B form the compounds

$B_2A_3$ and

$B_2A$ . If 0.05 mole of

$B_2A_3$ weighs 9 g and 0.1 mole of

$B_2A$ weighs 10 g, the atomic weight of A and B respectively are:

0%
30 and 40
0%
40 and 30
0%
20 and 5
0%
15 and 20

Explanation

$moles =\dfrac{ mass}{molar\, mass}$

then,

$molar \, mass = \dfrac{mass}{moles}$

$B_{ 2 }A_{ 3 }$ molar mass

$={ 9 }{ 0.05 }=180$

relative formula mass of

$B_{2}A_{3}=180$

that means,

(atomic mass of

$A$ )

$\times 3 +$ (atomic mass of

$B$ )

$2=180$

$3A+2B=180$ .....(i)

$B_{2}A$ molar mass

$=\dfrac{10}{0.1} = 100u$

$2B+A=100$ .....(ii)

from equation (i) and (ii)

$A=40$

$B=30$

Hence, the correct option is

$B$

The mass of an element

$E$ found in

$1.00\ mole$ of each of four different compounds is

$38.0\ g$ ,

$57.0\ g$ ,

$76.0\ g$ and

$114\ g$ respectively .

What is possible atomic weight of

$E$ ?

0%
$19.0$
0%
$38$
0%
$57$
0%
$76$

Explanation

mass of an element in a different compound is

$38\ g, 31\ g, 114\ gm$

$HCF$ of all there may be the possible weight of

$E$

So, the atomic weight is

$19.0$

Energy of the third orbit of Bohr's atom is:

0%
$-13.6$ eV
0%
$-3.4$ eV
0%
$-1.5$ eV
0%
none of these

Explanation

$E_n = - 13.6\times(z^2/n^2)eV$

$E_3 = - 13.6\times1/9eV$

$= - 1.5eV$

Which element with lowest atomic number .

0%
$_{19}K$
0%
$_{24}Cr$
0%
$_{11}Na$
0%
$_{29}Cu$

Explanation

The atomic number is defined as the number of protons present in an element.

In sodium atom, the number of protons present =

$11$

So, its atomic number is

$11$

Which is least from the rest of the given elements.

Hence, the correct option is

$C$

What is the ratio of time periods

$(\dfrac{T_1}{T_2})$ in

$2^{nd}$ orbit of hydrogen atom to third orbit of

$He^+$ ion?

0%
$\dfrac{8}{27}$
0%
$\dfrac{32}{27}$
0%
$\dfrac{27}{32}$
0%
$\dfrac{27}{8}$

Explanation

$\dfrac{T_H}{T_{He}}=\dfrac{n_1^3t_2^2}{n_2^3t_1^2}$

$= \dfrac{8\times4}{3^3}=\dfrac{32}{27}$

The ratio is

$32:27$ .

If the ratio of the radius of the two different nuclei is 2:3; then what will be the ratio of their mass?

(If density is same)

0%
16:27
0%
4:81
0%
8:27
0%
5:25

Explanation

$r_1 = 2, r_2 = 3$

$d =$ density

$=$ mass /volume

$\dfrac{d_1}{d_2} = \dfrac{m_1/(4/3\pi r_1^3)}{m_2/(4/3\pi r_2^3)}$

Ratio of densities is 1 because they are same

$\dfrac{m_1}{m_2} =\dfrac{ r_2^3}{r_1^3} = \dfrac{27}{8}$

Or ratio

$\dfrac{m_2}{m_1} = \dfrac{8}{27}$

A single electron orbits a stationary nucleus of charge

$+Z$ , where

$Z$ is a constant. It requires 47.2 eV to excite electron from second Bohr orbit to third Bohr orbit, find the value of

$Z$ .

0%
1
0%
3
0%
5
0%
4

Explanation

Position is

$n_1=2 \ \Rightarrow n_2=3$ and

$\Delta E=47.2eV$

We have

$\Delta E=13.6z^2(\dfrac{1}{n_1^2}− \dfrac{1}{n_2^2})eV$

$47.2=13.6z^2(\dfrac{1}{2^2}− \dfrac{1}{3^2})$

$z=5$

Multiple of fine structure of spectral lines is due to:

0%
presence of main energy levels
0%
presence of sub-level
0%
presence of electronic configuration
0%
is not a characteristics of the atom.

Explanation

It is because of sub energy levels. Previously the energy of sub energy levels was considered to be equal but it actually differs by about

$10^{-4}$ eV. Hence fine spectral lines are observed

If the speed of electrons in Bohr's first obtain of hydrogen is

$'X'$ the speed of electrons in third orbit is:

0%
$\dfrac {X}{9}$
0%
$\dfrac {X}{3}$
0%
$3X$
0%
$9X$

Explanation

We know,

$v = v_0 z/n$ here

$v_0 = X =$ velocity of elctron in first orbit of hydrogen or velocity constant

for third orbit of hydrogen atom value of n is 3 and z is 1

$v = \dfrac{X }{3}$

For Hydrogen atom the ratio of time required to complete

$1$ revolution of electron in

$I^{st},II^{nd}$ and

$III^{rd}$ orbit respectively is:

0%
$1:8:27$
0%
$1:4:9$
0%
$1:2:3$
0%
$1:16:81$

Explanation

Time required to complete

$1$ revolution

$=\cfrac{2\pi r}{v}$

$\because r\propto n^2$ and

$v\propto \cfrac{1}{n}$

$\therefore$

$Time$

$required\propto n^3$

$\therefore$ Ratio of time required in

$I^{st},II^{nd}$ and

$III^{rd}$ orbit is,

$(1)^3:(2)^3:(3)^3$ i.e.

$1:8:27$

What happens to the atomic number during a chemical reaction?

0%
It increases.
0%
It changes.
0%
Remains the same.
0%
Changes alternatively.

Which of the following parameters are not same for all hydrogen like atoms and ions in their state?

0%
Radius of orbit
0%
Speed of electron
0%
Energy of the atom
0%
Orbital angular momentum of electron

Explanation

As we know,

$r \propto n^2$

$E \propto -\dfrac{z^2}{n^2}$

$V\propto \dfrac{z}{n}$

$mur=nh^2 \pi$

So, independent of $z$ so it will remain same.

So, orbital angular momentum of electron

Which of the following frequencies is associated with the visible region of a spectrum?

0%
$3\times10^9$ cycle /seconds
0%
$6\times10^{14}$ cm cycle /seconds
0%
$5.0\times10^{10-3}$ cycle /seconds
0%
None of these

Explanation

Frequency range of visible region

$430-770$

$TH_{2}$

$430\times 10^{12}-770\times 10^{12}Hz$

$6\times 10^{14}$ cycle/second associated with visible region

The ratio of the radius of

$1st$ orbit of

$H$ atom to that of its

$4$ th excited state is:

0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{25}$
0%
$25$
0%
$\dfrac{1}{16}$

Explanation

Bohr radius r

$\propto \dfrac{n^{2}}{2}$

$\dfrac{r_{1st} \, orbit }{r _{4th} \, excited \, state} = \dfrac{1^{2}}{4^{2}}$

$\dfrac{1}{16}$

Option D is the correct answer.

According to Bohr's atomic theory, which of the following is correct?

0%
Potential energy of electrom $\alpha$ $\dfrac{Z^2}{n^2}$
0%
The product of velocity of electron and principle quantum number (n) $\alpha Z^2$
0%
Frequency of revolution of electron in an orbit $\alpha \dfrac{Z^2}{n^3}$
0%
Coulombic force of attraction on the electron $\alpha \dfrac{Z^2}{n^2}$

Explanation

Potential energy of an electron is given by,

$P.E=-\dfrac{2\pi^2 mZ^2 e^4}{n^2h^2}$

Therefore,

$P.E$

$\alpha$

$\dfrac{Z^2}{n^2}$

velocity of an electron is given by,

$v=\dfrac{2\pi Ze^2}{nh}$

Therefore,

$(v\times n)$

$\alpha$

$Z$

Frequency of an electron is given by,

$f=\dfrac{4\pi^2 Z^2 e^4}{n^3 h^3}$

$f$

$\alpha$

$\dfrac{Z^2}{n^3}$

Coloumbic force of attraction is given by,

Coloumbic force of attraction of an electron

$\alpha$

$\dfrac{Z^3}{n^4}$

A trend common to both group

$I$ and

$VII$ elements in the periodic table as atomic number increase is

0%
Atomic radius increase
0%
Oxidising power increase
0%
Reactivity with water increase
0%
Maximum valency increase

Potassium selenate is isomorphous with the potassium sulphate and contains 50.0% of

$Se$ . Find the atomic weight of

$Se$ .

0%
142
0%
71
0%
47.33
0%
284

Explanation

Pottassium selenate is isomorphous to $K_2SO_4$ and thus its molecular formula is $K_2SeO_4$

Now mol.wt of $K_2SeO_4=(39\times 2+a+4\times 16)$

$\Rightarrow 142+a$

Where $'a'$ is at.wt of $Se$

$(142+a)g\;K_2SeO_4$ $=Se\ a\ g$

100g $K_2SeO_4=\dfrac{a\times 100}{142+a}$

$\%$ of $Se=50$

$a=142$

$\approx 142$

Also equivalent weight of $K_2SeO_4=\dfrac{Mol.wt}{2}$

$\Rightarrow \dfrac{2\times 39+142+64}{2}$

$\Rightarrow 142$

Hence $(A)$ is the correct answer.

Atomic number of the central atom in

${ MCl }_{ 2 }$ is

$50$ . The shape of gaseous

${ MCl }_{ 2 }$ is given as:

Explanation

$MC{{L}_{2}}$ have $50$ atomic number so $M$ will be $16$ atomic number.

$C{{l}_{2}}=17+17$

atomic number is $34+16=50$ .

And molecule will be $SC{{l}_{2}}$ and according to this shape of the molecule is bent.

And $S$ have two lone pair.

The moment of momentum of electrons in the third orbit in a hydrogen atom

0%
$31.67 \times 10^{-34} gm \ m^2/s$
0%
$3.167 \times 10^{-34} kg \ m^2/s$
0%
$31.67 \times 10^{-34} kg \ cm^2/s$
0%
$31.67 \times 10^{-34} gm \ cm^2/s$

Ionization energy of

${He}^{+}$ is

$19.6\times {10}^{-18}J{atom}^{-1}$ . The energy of the first stationary state (

$n=1$ ) of

${Li}^{2+}$ is

0%
$4.41\times {10}^{-16}J{atom}^{-1}$
0%
$-4.41\times {10}^{-17}J{atom}^{-1}$
0%
$-2.2\times {10}^{-15}J{atom}^{-1}$
0%
$-8.82\times {10}^{-17}J{atom}^{-1}$

Explanation

$IE$ of

${ He }^{ + }$ =

$19.6\times { 10 }^{ -18 } J$ per atom

$IE\quad =\quad { E }_{ 1\quad }(for\quad H)\times { Z }^{ 2 }$

${ E }_{ 1 }\times 4= -19.6\times { 10 }^{ -18 }\quad J$

${ E }_{ 1 }(for \; {Li}^{+2})={ E }_{ 1 }(for\quad H)\times 9$

$- \cfrac { 19.6\times { 10 }^{ -18 } }{ 4 } \times 9$

$- 441\times { 10 }^{ -18 }$

$- 4.41\times { 10 }^{ -17 } J$ per atom

Approximate atomic weight is of an element is

$26.89 \; g$ . If its equivalent weight is

$8.9 \; g$ , the exact atomic weight of element would be:

0%
$29.89$
0%
$8.9$
0%
$17.8$
0%
$26.7$

Explanation

Atomic weight = Equivalent weight

$\times$ Valency

$\Rightarrow$ Valency

$= \cfrac{\text{approx. at. wt.}}{\text{Eq. wt.}}$

Eq. wt.

$= 8.9 g \; \left( \text{Given} \right)$

$\therefore$ Valency

$= \dfrac{26.89}{8.9}=3$

$\Rightarrow$ At. wt.

$= 8.9 \times 3= 26.7 \; g$

So, the correct option is

$D$

Name the compounds used in preparation of nylon-66.

0%
E-caprolactum
0%
Hexamethylenediamine and adipic acid
0%
Dimethyl tetraphthalate
0%
Hexamethylene diamine

The radius of first excited state of

${ Be }^{ 3+ }$ ion is:

0%
$0.53\overset { o }{ A }$
0%
$\cfrac { 0.53 }{ 4 } \overset { o }{ A }$
0%
$0.53\times 4\overset { o }{ A }$
0%
$0.53\times 2\overset { o }{ A }$

The e/m of proton is

0%
$1.78\times {10}^{8}C/g$
0%
$9.57\times {10}^{4}C/g$
0%
$19.14\times {10}^{4}C/g$
0%
$0.478\times {10}^{4}C/g$

Which of the following hydrogen bonds are strongest in vapour phase?

0%
$HF.....HF$
0%
$HF...HCl$
0%
$HCl...HCl$
0%
$HF...HI$

The unfavorable electrochemical reaction among the following is:

The radius of the hydrogen atom in the ground state is

$0.53 {\mathring A}$ , the radius of

$L{i^{2 + }}$ in a similar state is____.

0%
$1.06 {\mathring A}$
0%
$0.265 {\mathring A}$
0%
$0.175 {\mathring A}$
0%
$0.53 {\mathring A}$

Explanation

Radius of hydrogen like particle

$r_n=a_0\dfrac{n^2}{2}$

$0.530 \mathring { A }=a_0\dfrac{1^2}{1}$

$a_0=0.53\mathring{A}$

for

$Li^{2+}$

$r=a_0\dfrac{1^2}{3}$

$=\dfrac{0.530}{3}=0.176 \mathring{A}$

Hence, option

$C$ is correct.

Which among the following elements does not contain same number of s-electrons as the number of d-electrons in

${ Ni }^{ 2+ }$ ? (Atomic number of Kr is 36)

0%
Ca
0%
Kr
0%
Cu
0%
Fe

Which of the atomic number represents s -block elements?

0%
7, 15
0%
3, 12
0%
6, 12
0%
9, 17

Explanation

The elements in which the last electron enters the outermost s-orbital are called s-block elements.

The s-block elements have two groups ( $1$ and $2$ ).

Thus, s-block elements include lithium $(Li)$ and magnesium $(Mg).$

The periodic table shows exactly where these elements are within the s-block.

The radius of the first Bohr orbit of a hydrogen atom is

$0.53\times 10^{-10}m$ .When an electron collides with this atom which is in its normal state, the radius of the electron orbit in the atom change to

$2.12 \times 10^{10}m$ .The value of the principal quantum number

$n$ of the state to which it is excited is:

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

$r_n=a_0n^2$

$2.12\times 10^{-19}=90n^2$

$=0.53n^2$

$n^2=4$

$n=2$
This value of principle quantum number of excited state in

$2$

If transition of an electron in

$H-atom$ takes place from

$3rd$ orbit to

$5th$ then change in angular momentum of electron is given by?

0%
$h$
0%
$\dfrac{h}{2}$
0%
$2h$
0%
$3h$

Bohr's theory can be applied to determine the?

0%
electron gain enthalpy of hydrogen atom
0%
Electronegativity of ${ Li }^{ 2+ }$ ion
0%
second ionization energy of helium atom
0%
first ionization energy of Be atom

The ratio between the number of neutrons present in

$C^{12}$ and

$Si^{30}$ atoms is:

0%
3:8
0%
2:5
0%
3:7
0%
1:1

The ratio of the energy of the electron in the ground state of hydrogen to the electron in first excited state of

$Be^{3+}$ is

0%
$1 : 4$
0%
$1 : 8$
0%
$1 : 16$
0%
$16 : 1$

Explanation

Formula,

$E=E_0 \dfrac{z^2}{n^2}$

hydrogen,

$E=E_0 \dfrac{z^2}{n^2}=E_0 \dfrac{1^2}{1^2}$

$E=E_0$

$Be^{3+},$

$E=E_0 \dfrac{z^2}{n^2}=E_0 \dfrac{4^2}{2^2}$

$E=4E_0$

$Ratio=\dfrac{E_0}{4E_0}$

$1:4$

In a hydrogen atom the force acting on the electron is proportional to :

0%
$n^{4}$
0%
$n^{-4}$
0%
$r^{2}$
0%
$n^{-2}$

Explanation

A hydrogen atom the force acting on the election is proportional to

$r^2$ .

Hence ,option

$C$ is correct.

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 9 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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