Structure of Atom Test

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Structure of Atom Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

A certain sample of element Z contains 60% of $${ 69 }_{ Z }$$ and 40% of $$71_{ Z }$$. What is the relative atomic mass of element Z in this sample.

A
69.2

B
69.8

C
70.0

D
70.2

Solution

RAM=$$\dfrac{60\times 69+71\times 40}{100}$$

$$=\dfrac{414+284}{10}$$

$$=\dfrac{698}{10}=69.84$$
Question 2
1 / -0

Which of the following statement is not correct?

A
Atomic no.of A,B,C are $$40,57,105$$ respectively

B
Group no of A, B,C are IV B, III B and V B respectively

C
Period no of A, B,C are 4th , 5th and 6th respectively

D
C is a radioactive element
Solution
(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 $$A\longrightarrow 40, B\longrightarrow 57, C\longrightarrow 105$$
Question 3
1 / -0

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?

A
21

B
11

C
401

D
19

Solution

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 $$21$$.
Question 4
1 / -0

Visible spectrum lies between:

A
400nm to 760 nm

B
350nm to 760 nm

C
400nm to 660 nm

D
300nm to 760 nm

Solution

Visible Spectrum lies between 400nm to 760 nm.
Question 5
1 / -0

What is the potential energy of an electron present in N-shell of the $${ B }e^{ 3+ } ion$$?

A
$$-3.4 eV$$

B
$$-6.8 eV$$

C
$$-13.6 eV$$

D
$$-27.2 eV$$

Solution

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 $$D$$
Question 6
1 / -0

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

A
$$1$$st orbit

B
$$2$$nd orbit

C
$$7$$th orbit

D
infinite orbit

Solution

When electron is in the first orbit, it has the lowest energy, but when it absorbs energy, it enters into second, third orbits based on energy. It never emits energy as it has lowest energy.
Question 7
1 / -0

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 $$E$$?

A
$$19.0$$

B
$$38$$

C
$$57$$

D
$$76$$

Solution

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 $$19.0$$
Question 8
1 / -0

$$4.16 g$$ of $$BaCI_{2}$$ contains $$2.74 g$$ of Ba. The atomic weight of Ba is nearly:

A
$$137$$

B
$$208$$

C
$$67.3$$

D
$$97.27$$

Solution

$$Ba^{2+}+2Cl^-\longrightarrow BaCl_2$$
Weight of $$Ba=2.74g$$
Weight of$$ BaCl_2=4.16g$$
Formula :
$$\cfrac{\text{Moles of }Ba^{2+}}{\text{Stoichiometry of } Ba^{2+}}=\cfrac{\text{Moles of }BaCl_2}{\text{stoichiometry of }BaCl_2}$$
$$\therefore \cfrac{\text{Moles of }Ba^{2+}}{1}=\cfrac{\text{Moles of }BaCl_2}{1}$$
$$\therefore\cfrac{weight}{\text{GMM of } Ba^{2+}}=\cfrac{weight}{\text{GMM of } BaCl_2}$$
$$\cfrac { 2.74g }{ { GMMof }Ba^{ 2+ } } =\cfrac { 4.16g }{ { 208.23\quad gmol^{ { -1 } } } } $$
GMM of $$Ba^{2+}=\cfrac{2.74g\times 208gmol^{-1}}{4.16g}$$
Atomic mas of $$Ba=137.15$$
Question 9
1 / -0

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)

A
16:27

B
4:81

C
8:27

D
5:25

Solution

$$r_1 = 2, r_2 = 3$$
$$d =$$ 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 $$ \dfrac{m_2}{m_1} = \dfrac{8}{27}$$
Question 10
1 / -0

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:

A
30 and 40

B
40 and 30

C
20 and 5

D
15 and 20

Solution

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

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 $$B$$