States of Matter Test

Result

States of Matter Test - 50

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

The intermolecular forces of attraction (i.e., H-bonding and van der Waals' forces) exist among polar and non-polar species which affect melting point, boiling point, solubility and viscosity of covalent compounds :

...view full instructions

The type of molecular force of attraction present in the following compound is :

A
Intermolecular H-bonding

B
Intramolecular H-bonding

C
van der WaaLs' force

D
All of these

Solution
Question 2
1 / -0

Directions For Questions

The intermolecular forces of attraction (i.e., H-bonding and van der Waals' forces) exist among polar and non-polar species which affect melting point, boiling point, solubility and viscosity of covalent compounds :

...view full instructions

Select the incorrect order of boiling point between the following compounds :

A
$$N_3H < CH_3N_3$$

B
$$Me_2SO_4 < H_2SO_4$$

C
$$Me_3BO_3 < B(OH)_3$$

D
$$BF_3 < BI_3$$

Solution

$$\rightarrow$$ Boiling point of $$HN_3 > CH_3N_3$$ as in hydrazoic acid ($$HN_3$$) intermolecular H-bonding occurs and its bond energy is higher than weak dipole-dipole interactions present in $$CH_3N_3$$.
$$\rightarrow$$ Boiling point of $$BI_3$$ is greater than that of $$BF_3$$ as molecular weight of $$BI_3$$ is higher than that of $$BF_3$$.
$$\rightarrow$$ Due to intermolecular H-bonding in $$H_2SO_4$$ its boiling point is higher than $$Me_2SO_4$$.
$$\rightarrow$$ Similarly due to intermolecular H-bonding in $$B(OH)_3$$ its boiling point is higher. than that of $$Me_3BO_3$$.
Question 3
1 / -0

At a certain pressure, volume of the gas at $$27^\circ C$$ is $$20$$ litre. If the pressure and temperature are doubled, the volume will be

A
$$40\,litre$$

B
$$20\,litre$$

C
$$10.9\,litre$$

D
$$8.2\,litre$$

Solution

According to question:

$$P_1 =P$$, $$V_1 =20L$$, $$T_1 =T$$

$$P_2=2P$$, $$V_2=?$$, $$T_2= 2T$$

Now,

$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$

$$\dfrac{P \times 20}{T}=\dfrac{2P \times V_2}{2T}$$

$$\Rightarrow V_2=20\,L$$
Question 4
1 / -0

Kinetic theory of gases presumes the collision between the molecules to be perfectly elastic because:

A
the molecules are rigid

B
the temperature remains constant irrespective of collisions

C
collisions will not split the molecules

D
the molecules are tiny

Solution

The kinetic theory of gases presumes the collision between the molecules to be perfectly elastic because the molecules are tiny. If the molecules are tiny, the energy and momenta of the molecules are conserved during the collisions. Consequently, the temperature remains constant irrespective of collisions.
Question 5
1 / -0

Which of the given laws of chemical combination is satisfied by the figure?

A
Law of multiple proportion

B
Gay Lussac's law

C
Avogadro law

D
Law of definite proportion

Solution
Question 6
1 / -0

"Equal volumes of all gases at the same temperature and pressure contain equal number of particles." This statement is a direct consequence of:

A
Avogadro's law

B
Charles law

C
Ideal gas equation

D
Law of partial pressure

Solution

Avogadro's Law states that "Equal volumes of all gases at the same temperature and pressure contain equal number of particles."

Hence, Option "A" is the correct answer.

Question 7

1 / -0

Match the following :

List -I	List-II
A. Boyle’s law	1. $$P_{wet}=P_{dry}+P_{water\ vapour}$$
B. Avogadro law	2. $$V_{1}=n_{1}\left ( \dfrac{V_{2}}{n_{2}} \right )$$
C. Charles law	3. $$V_{1}=V_{0}\left (1+ \dfrac{t}{273} \right )$$
D. Dalton’s law	4. $$V_{1}=P_{2}\left ( \dfrac{V_{2}}{P_{1}} \right )$$

The correct match is:

A-1, B-2, C-3, D-4

A-2, B-3, C-4, D-1

A-4, B-3, C-2, D-1

A-4, B-2, C-3, D-1

Solution

Boyle's law: At constant temperature, $$PV=$$ constant

$$\therefore P_1V_1 = P_2V_2$$

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

$$\therefore \dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}$$

Charles law states that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature if the pressure remains constant.

$$= V_{1}=V_{0}\left (1+ \dfrac{t}{273} \right )$$

Dalton's law: Net vapour pressure is the addition of vapour pressure of individual components.

$$\therefore P_{wet} = P_{dry} + P_{water\ vapour}$$

Hence, the correct answer is option $$\text{D}$$.

Question 8
1 / -0

2.5 L of a sample of a gas at $$27^\circ C$$ and 1 bar pressure is compressed to a volume of 500 mL keeping the temperature constant, the percentage increase in the pressure is :

A
100%

B
400%

C
500%

D
80%

Solution
Question 9
1 / -0

Dalton's law of partial pressures is not applicable to which one of the following?

A
$$H_{2}+Cl_{2}$$

B
$$SO_{2}+Cl_{2}$$

C
$$NH_{3}+HCl$$

D
All the above

Solution

According to Dalton's law of partial pressures, for a mixture of $$\text{non-reacting}$$ gases, the total pressure is equal to the sum of the partial pressures of the individual gases.

This law is not applicable to the mixtures $$H_{2}+Cl_{2}, NH_{3}+HCl $$ and $$SO_{2}+Cl_{2}$$ as they react with each other.

For example, hydrogen and chlorine react to form hydrogen chloride, ammonia and hydrochloric acid forms ammonium chloride while sulphur dioxide reacts with chlorine to form $$SO_2Cl_2$$.

Hence, the correct answer is option $$\text{D}$$.
Question 10
1 / -0

At the same temperature, HCl gas and NH$$_{3}$$ gas are present in two vessels of same volume at a pressure of 'P' atmospheres each. When one jar is inverted over the other so that the two will mix, after some time the pressure in the vessels will become:

A
$$\cfrac{P}{2}$$

B
$$\cfrac{P}{4}$$

C
Zero

D
P

Solution

$$NH_3 + HCl \rightarrow NH_4Cl(solid)$$

Due to the formation of a solid, final pressure in the vessel will become zero.

Hence, the correct answer is option $$\text{C}$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

States of Matter Test - 50

Result

States of Matter Test - 50

Score

-

Rank

-

SHARING IS CARING

Question 1

Intermolecular H-bonding

Intramolecular H-bonding

van der WaaLs' force

All of these

Solution

Question 2

$$N_3H < CH_3N_3$$

$$Me_2SO_4 < H_2SO_4$$

$$Me_3BO_3 < B(OH)_3$$

$$BF_3 < BI_3$$

Solution

Question 3

$$40\,litre$$

$$20\,litre$$

$$10.9\,litre$$

$$8.2\,litre$$

Solution

Question 4

the molecules are rigid

the temperature remains constant irrespective of collisions

collisions will not split the molecules

the molecules are tiny

Solution

Question 5

Law of multiple proportion

Gay Lussac's law

Avogadro law

Law of definite proportion

Solution

Question 6

Avogadro's law

Charles law

Ideal gas equation

Law of partial pressure

Solution

Question 7

A-1, B-2, C-3, D-4

A-2, B-3, C-4, D-1

A-4, B-3, C-2, D-1

A-4, B-2, C-3, D-1

Solution

Question 8

100%

400%

500%

80%

Solution

Question 9

$$H_{2}+Cl_{2}$$

$$SO_{2}+Cl_{2}$$

$$NH_{3}+HCl$$

All the above

Solution

Question 10

$$\cfrac{P}{2}$$

$$\cfrac{P}{4}$$

Zero

P

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.