States of Matter Test

Result

States of Matter Test - 23

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

An aqueous solution of methanol in water has vapour pressure:

A
less than that of water

B
equal to that of water

C
more than that of water

D
equal to that of methanol

Solution

Correct Answer: Option (c)

Hint: $$A-A$$ and $$B-B$$ interactions are greater than $$A-B$$ interaction in positive deviation from Raoult's law

Explanation:

· Methanol and hydrogen have strong hydrogen bonding between them Methanol and water form hydrogen bonds between them. The hydrogen bonding is stronger than that between methanol molecules but the interactions are not as extensive as in water.

· Thus, it has a vapour pressure lower than methanol and higher than water. Overall $$A - A$$ and $$B - B$$ interaction $$ > A - B$$ interactions and the solution show positive deviation.

Final answer: An aqueous solution of methanol in water has vapour pressure more than that of water so the correct answer is (c)
Question 2
1 / -0

What is the vapour pressure of a solution of glucose which has an osmotic pressure of 3 atmospheres at $$20^C?$$ The vapour pressure of water at $$20^oC$$ is $$17.39 mm$$. Consider the density of solution equal to that of the solvent.

A
$$12.35 mm$$

B
$$14.35 mm$$

C
$$16.35 mm$$

D
$$17.35 mm$$

Solution
Question 3
1 / -0

At $$25^oC$$, the vapour pressure of pure benzene is 100 torr, while that of pure ethyl alcohol is 44 torr. Assuming ideal behaviour, calculate the vapour pressure at $$25^oC$$ of a solution which contains 10 g of each substance.

A
33.775 torr

B
54.775 torr

C
64.775 torr

D
60.775 torr

Solution

We have,
moles of benzene = $$\dfrac{10}{78}$$ = 0.128
moles of ethyl alcohol = $$\dfrac{10}{46}$$ = 0.217

$$P_t = P^0_A * x_A + P^0_B * x_B$$

$$P_t = 100 * \dfrac{0.128}{0.345} + 44*\dfrac{0.217}{.345}$$

$$P_t = 64.77 torr$$
Question 4
1 / -0

Directions For Questions

Kinetic theory of gases is a generalization offered by Maxwell Boltzmann, Clausius, etc., to explain the behaviour of ideal gases. This theory assumes that ideal gas molecules neither attract nor repel each other. Average kinetic energy of gas molecules is directly proportional to the absolute temperature. A gas equation called kinetic gas equation was derived on the basis of kinetic theory.
$$PV = \dfrac {1}{3}mnv^{2}$$.

...view full instructions

Which of the following do not pertain to the postulates of kinetic theory of gases?

A
The gas molecules are perfectly elastic

B
Speed of gas molecules are every changing

C
Pressure exerted by the gas is due to the collision of molecules with the walls of the container

D
Kinetic energy of a gas is given by the sum of $$273$$ and temperature in Celsius scale

Solution

The option (B) do not pertain to the postulates of kinetic theory of gases.

"Speed of gas molecules are every changing."

All that is mentioned in the postulates of kinetic theory of gases is that the molecules are moving very fast in all directions in a straight line until their direction is changed on collision with each other or with the walls of containers. During collisions, there is redistribution of kinetic energy. Nothing is mentioned regarding ever changing speed.
Note:
Kinetic energy of a gas is directly proportional to the absolute temperature which is given by the sum of 273 and temperature in Celsius scale.
Question 5
1 / -0

Plot of Maxwell's distribution of velocities is given below:
which of following is correct?

A
$$T_{1} > T_{2}$$

B
$$T_{1} < T_{2}$$

C
$$v_{1} < v_{2}$$

D
$$f_{1} > f_{2}$$

Solution

On increasing temperature, the molecular velocity increases but fraction of the molecules possessing that velocity decreases. Therefore, $$T_1>T_2$$.
Question 6
1 / -0

Which of the following property indicates weak intermolecular forces of attraction in liquid?

A
High heat of vaporization

B
High vapour pressure

C
High critical temperature

D
High boiling point

Solution

The molecules that are having weak intermolecular forces of attraction, they can easily dissociate and convert into vapour state,
so, molecules with weak intractions have high vapour pressure.
Hence, option $$(B)$$ is correct.
Question 7
1 / -0

At $$27^{\circ}C$$ a sample of ammonia gas exerts a pressure of $$5.3\ atm$$. What is the pressure when the volume of the gas is reduced to one-tenth of the original value at the same temperature?

A
$$0.53\ atm$$

B
$$5.3\ atm$$

C
$$53\ atm$$

D
None of these

Solution

$$\begin{array}{l} Pv=nRT \\ { p_{ 1 } }{ v_{ 1 } }={ p_{ 2 } }{ v_{ 2 } } \\ \Rightarrow 5.3\lambda \frac { v }{ 1 } ={ p_{ 2 } }\times \frac { { \frac { v }{ 1 } } }{ { 10 } } \\ \Rightarrow { p_{ 2 } }=53\, atm. \\ Hence,\, the\, option\, C\, is\, the\, required\, answer. \end{array}$$
Question 8
1 / -0

A vessel of volume $$5$$ liter contains $$1.4\ g$$ of nitrogen at a temperature $$1800\ K$$. The pressure of the gas if $$30$$% of its molecules are dissociated into atoms at this temperature is :

A
$$4.05\ atm$$

B
$$2.0245\ atm$$

C
$$3.84\ atm$$

D
$$1.92\ atm$$

Solution

Reaction involved, dissociation of nitrogen gas
$$N_2\rightleftharpoons 2N$$
mass of nitrogen $$=1.4\,g$$
$$\Rightarrow moles =1.4/28=0.05$$
moles after dissociation of $$N_2=1.05\times \dfrac{70}{100}$$
and after dissociation of $$2N=0.05\times 2\times \dfrac{30}{100}$$
total moles involved $$=0.065$$
applying ideal gas equation $$(PV=nRT)$$
$$P=\dfrac{(nRT)}{V}$$
$$=1.92\,atm$$
Question 9
1 / -0

The temperature, at which the density of $$O_{2}$$ at $$1\ atm$$, is the same as that of $$CH_{4}$$ at S.T.P is:

A
$$273^{\circ}C$$

B
$$100^{\circ}C$$

C
$$546^{\circ}C$$

D
$$150^{\circ}C$$

Solution

$$\begin{array}{l} density\alpha \frac { { pressure } }{ { \sqrt { temperature } } } \\ \frac { { { d_{ { o_{ 2 } } } } } }{ { { d_{ C{ H_{ 4 } } } } } } =\frac { { P{ o_{ 2 } }\times \sqrt { { T_{ C{ H_{ 4 } } } } } } }{ { \sqrt { { T_{ { o_{ 2 } } } } } \times { P_{ C{ H_{ 4 } } } } } } \\ \sqrt { { T_{ { o_{ 2 } } } } } =\frac { { 1\times \sqrt { 373 } } }{ { 0.99 } } \\ { T_{ { o_{ 2 } } } }=\frac { { 373 } }{ { { { \left( { 0.99 } \right) }^{ 2 } } } } =380.57=\sim { 100^{ \circ } }C \\ Hence,\, option\, B\, \, is\, the\, correct\, answer. \end{array}$$
Question 10
1 / -0

Equal weights of two gases of molecular weight $$4$$ and $$40$$ are mixed. The pressure of the mixture is $$1.1$$ atm. The partial pressure of the light gas in this mixture is :

A
$$0.55$$ atm

B
$$0.15$$ atm

C
$$1$$ atm

D
$$0.11$$ atm

Solution

Number of moles of lighter gas $$=\dfrac{m}{4}$$

Number of moles of heavier gas $$=\dfrac{m}{40}$$

Total Number of moles $$=\dfrac{m}{4} + \dfrac{m}{40} = \dfrac{11m}{40}$$

Mole fraction of lighter gas $$=\dfrac{m}{40} \times \dfrac{40}{11m} = \dfrac{10}{11} $$

Partial pressure due to lighter gas $$= P_o \times \dfrac{10}{11}$$

$$=1.1 \times \dfrac{10}{11} = 1 atm$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

States of Matter Test - 23

Result

States of Matter Test - 23

Score

-

Rank

-

SHARING IS CARING

Question 1

less than that of water

equal to that of water

more than that of water

equal to that of methanol

Solution

Question 2

$$12.35 mm$$

$$14.35 mm$$

$$16.35 mm$$

$$17.35 mm$$

Solution

Question 3

33.775 torr

54.775 torr

64.775 torr

60.775 torr

Solution

Question 4

The gas molecules are perfectly elastic

Speed of gas molecules are every changing

Pressure exerted by the gas is due to the collision of molecules with the walls of the container

Kinetic energy of a gas is given by the sum of $$273$$ and temperature in Celsius scale

Solution

Question 5

$$T_{1} > T_{2}$$

$$T_{1} < T_{2}$$

$$v_{1} < v_{2}$$

$$f_{1} > f_{2}$$

Solution

Question 6

High heat of vaporization

High vapour pressure

High critical temperature

High boiling point

Solution

Question 7

$$0.53\ atm$$

$$5.3\ atm$$

$$53\ atm$$

None of these

Solution

Question 8

$$4.05\ atm$$

$$2.0245\ atm$$

$$3.84\ atm$$

$$1.92\ atm$$

Solution

Question 9

$$273^{\circ}C$$

$$100^{\circ}C$$

$$546^{\circ}C$$

$$150^{\circ}C$$

Solution

Question 10

$$0.55$$ atm

$$0.15$$ atm

$$1$$ atm

$$0.11$$ atm

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.