Atoms and Molecules Test

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Atoms and Molecules Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

Molecular mass of ${ Na }_{ 2 }{ SO }_{ 4 }$ . $10{ H }_{ 2 }O$ is:

A
320

B
321

C
322

D
324

Solution

$Na_2SO_4.10H_2O$
$=(2\times 23)+32+(4 \times 16)+(10\times 18)$
$=46+32+64+180$
$=322$
Question 2
1 / -0

Which of the following is the best example of law of conservation of mass?
[n and m are the masses of reactants and p and q are the masses of products formed.]

A
$n-m=p-q$

B
$n+m=p+q$

C
$n=m$

D
$p=q$

Solution

Law of conservation of mass:- Mass is neither created nor destroyed it just from one compound to other.

$n+m\rightarrow$ $p+q$

Hence, the mass of reactants is equal to the mass of products formed.
Option $B$ is correct.
Question 3
1 / -0

If Avogadro's number would have been $1\times { 10 }^{ 23}$ , instead of $6.02\times { 10 }^{ 23 }$ then mass of one atom of $_8^{16}O$ would be:

A
$1\ amu$

B
$1\times { 10 }^{ 10 }\ amu$

C
$16\ amu$

D
$6\times { 10 }^{ 13 }\ amu$

Solution

Atomic weight of $O=16\ g/mole$

We know that $1$ mole of $O$ contains $N_a=6.023\times 10^{23}$ number of atoms (Avogadro's Number of atoms)

Therefore, mass of $6.023\times 10^{23}$ atoms of $O=16\ g$

$\Rightarrow$ Mass of $1$ atom of $O=\dfrac{16\ g}{6.023\times 10^{23}}\ ....(i)$

Now, $1\ amu = \dfrac{1}{12}th$ the mass of one $^{12}C$ atom.

Atomic weight of carbon-12 atom $=12\ g$

$12\ g$ of $^{12}C$ atom contains $6.023\times 10^{23}$ atoms.

Therefore, mass of $1$ atom of $^{12}C=\dfrac{12}{6.023\times 10^{23}}$

$\therefore\ \dfrac{1}{12}th$ the mass of $1$ $^{12}C$ atom $=1\ amu= \dfrac{1}{12}\times \dfrac{12}{6.023\times 10^{23}}=\dfrac{1}{6.023\times 10^{23}}$

Now, according to the given condition, $N_a=1\times 10^{23}$ instead of
$6.023\times 10^{23}$ .

So, $1\ amu$ will now be $=\dfrac{1}{1\times 10^{23}}$

Now, putting the value of $1\ amu$ in equation $(i)$ , we get;

Mass of $1$ atom of $O=\dfrac{16}{1\times 10^{23}}\ g=16\times 10^{-23}\ g$

The value of mass of $1$ atom of oxygen in $amu$ :

${16}\times {10^{-23}}\times \dfrac{1}{1\times 10^{23}}\ amu$

$=16\ amu$

Hence, option (C) is correct.
Question 4
1 / -0

Assertion: For the formation of a molecule at least two atoms are needed.
Reason: Molecules are formed by bonding between different atoms of different elements or complexes without any exception.

A
Both Assertion and Reason are true and Reason is correct explanation

B
Both Assertion and Reason are true and Reason is not correct explanation

C
Assertion is correct but Reason is wrong

D
Both Assertion and Reason are wrong

Solution

A molecule forms when two or more atoms of the same elements join together by forming chemical bonds. So, the assertion is correct but the reason is not correct.
For example:
$Cl + Cl \rightarrow Cl_2$ (Chlorine molecule)
Hence, the correct option is C.
Question 5
1 / -0

How many moles of helium gas occupy $22.4L$ at $0^\circ C$ at $1$ atm pressure?

A
$0.11$ mole

B
$0.90$ mole

C
$1.0$ mole

D
$1.11$ mole

Solution

$pV=nRT$

$n=\dfrac { pV }{ RT }$

$=\dfrac { 1atm\left( 22.4L \right) }{ \left( 0.082Latm/molK \right) 273K }$

$n=1$ mole

$1$ mole of the gas occupies $22.4L$ at ${ O }^\circ C$ at $1atm$ pressure.

Hence, the correct option is C.
Question 6
1 / -0

Four different experiments were conducted in the following ways-

I) $3g$ of carbon was burnt in $8g$ of oxygen to give $11g$ of $CO_2$ .

II) $1.2g$ of carbon was burnt in air to give $4.2g$ of $CO_2$ .

III) $4.5g$ of carbon was burnt in enough air to give $11g$ of $CO_2$ .

IV) $4g$ of carbon was burnt in oxygen to form $30.3g$ of $CO_2$ .

Law of constant proportions is illustrated in which of the following experiment(s)?

A
I and III only

B
II and III only

C
IV only

D
I only

Solution

The law of definite proportion or law of constant composition states that a given chemical compound always contains its component elements in a fixed ratio (by mass) and does not depend on its source and method of preparation.

In the option, $A$ , $3g$ of carbon reacts with $8g$ of oxygen to form $11g$ of carbon dioxide.

$C + O_2 \xrightarrow[]{}CO_{2(g)}$
$3 + 8 = 11$

Thus, it follows the law of constant proportion. Option $D$ is correct.
Question 7
1 / -0

${ K }_{ 4 }\left[ Fe\left( CN \right) _{ 6 } \right]$ is supposed to be $40$ percent dissociated when $1M$ solution prepared. Its boiling point is equal to another $20$ percent mass by volume of non-electrolytic solution $A$ Considering molality=molarity . The molecular weight of $A$ is:

A
$77$

B
$67$

C
$57$

D
$47$

Solution
Question 8
1 / -0

The ratio of radius of the nucleus to the radius of the atom is of the order:

A
$10^5$

B
$10^6$

C
$10^{-5}$

D
$10^{-8}$

Solution

Nuclear radius $=10^{-15}$ m

Atomic radius $= 10^{-10}$ m
Ratio $= \dfrac{10^{-15}}{ 10^{-10}}=10^{-5}$
Hence, the correct option is $C$
Question 9
1 / -0

A mixture of $3\times10^{21}$ molecules of $X$ and $4.5\times10^{21}$ molecules of $Y$ weigh $1.375\ g$ . If the molecular weight of $X$ is $50\ g$ , then what is the molecular weight of $Y$ ?

A
$100$

B
$150$

C
$250$

D
$300$

Solution

Given,
Number of molecules of $X = 3\times 10^{21}$
Number of molecules of $Y = 4.5\times 10^{25}$
Molecular weight of $X = 50$
Total mass of mixture $= 1\ g$

Formula used :
$Mass\ of\ mixture=\left[\dfrac{Number\ of\ molecules\ of\ X}{Avogrado's\ Number} \times Molar\ Mass\ of\ X\right]$
$+\left[\dfrac{Number\ of\ molecules\ of\ Y}{Avogrado's\ Number}\times Molar\ Mass\ of\ Y\right]$

Now, putting all the given values in this formula, we get the molecular mass of $Y$ .
$1=\left[\dfrac{3\times 10^{21}}{6.023\times 10^{23}}\times 50\right]+\left[\dfrac{4.5\times 10^{25}}{6.023\times 10^{23}}\times Y\right]$

$\Rightarrow 1=\dfrac{150\times 10^{21}+4.5Y\times 10^{25}}{6.023\times 10^{23}}$

$\Rightarrow 6.023\times 10^{23}=10^{21}(150+4.5Y\times 10^4)$

$\Rightarrow \dfrac{6.023\times 10^{23}}{10^{21}}=150+4500Y$

$\Rightarrow 602.3=150+4500Y$

$\Rightarrow 4500Y=602.3-150$

$\Rightarrow Y = \dfrac{450.3}{4500}$

$Y= 150$

Therefore, on solving, we get $Y=150$
Therefore, the molecular weight of $Y$ is $150g$ .

Hence option (B) is correct.
Question 10
1 / -0

A box measures $10\ cm \times 11.2\ cm \times 10\ cm$ . Assume that this box is filled with neon gas at $1$ atm pressure and $273\ K$ temperature. How many electrons will be there in the box ?

A
$6.022\times 10^{23}$

B
$3.011\times 10^{23}$

C
$6.022\times 10^{22}$

D
$3.011\times 10^{22}$

Solution

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Atoms and Molecules Test - 61

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Atoms and Molecules Test - 61

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Question 1

320

321

322

324

Solution

Question 2

n−m=p−qn-m=p-qn−m=p−q

n+m=p+qn+m=p+qn+m=p+q

n=mn=mn=m

p=qp=qp=q

Solution

Question 3

1 amu1\ amu1 amu

1×1010 amu1\times { 10 }^{ 10 }\ amu1×1010 amu

16 amu16\ amu16 amu

6×1013 amu6\times { 10 }^{ 13 }\ amu6×1013 amu

Solution

Question 4

Both Assertion and Reason are true and Reason is correct explanation

Both Assertion and Reason are true and Reason is not correct explanation

Assertion is correct but Reason is wrong

Both Assertion and Reason are wrong

Solution

Question 5

0.110.110.11 mole

0.900.900.90 mole

1.01.01.0 mole

1.111.111.11 mole

Solution

Question 6

I and III only

II and III only

IV only

I only

Solution

Question 7

777777

676767

575757

474747

Solution

Question 8

10510^5105

10610^6106

10−510^{-5}10−5

10−810^{-8}10−8

Solution

Question 9

100100100

150150150

250250250

300300300

Solution

Question 10

6.022×10236.022\times 10^{23}6.022×1023

3.011×10233.011\times 10^{23}3.011×1023

6.022×10226.022\times 10^{22}6.022×1022

3.011×10223.011\times 10^{22}3.011×1022

Solution

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$n-m=p-q$

$n+m=p+q$

$n=m$

$p=q$

$1\ amu$

$1\times { 10 }^{ 10 }\ amu$

$16\ amu$

$6\times { 10 }^{ 13 }\ amu$

$0.11$ mole

$0.90$ mole

$1.0$ mole

$1.11$ mole

$77$

$67$

$57$

$47$

$10^5$

$10^6$

$10^{-5}$

$10^{-8}$

$100$

$150$

$250$

$300$

$6.022\times 10^{23}$

$3.011\times 10^{23}$

$6.022\times 10^{22}$

$3.011\times 10^{22}$