States of Matter Test

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States of Matter Test - 60

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Weekly Quiz Competition

Question 1
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In three beakers labeled as (A), (B) and (C), $$100mL$$ of water, $$100mL$$ of $$1M$$ solution of glucose in water and $$100mL$$ of $$0.5M$$ solution of glucose in water are taken respectively and kept at same temperature.
Which of the following statements is correct?

A
Vapour pressure in all the three beakers is same

B
Vapour pressure of beaker B is highest

C
Vapour pressure of beaker C is highest

D
Vapour pressure of beaker B is lower than that of C and vapour pressure of beaker C is lower than that of A

Solution

Liquid solution is formed when we dissolve a solid, liquid or gas in a particular liquid solvent. Vapour pressure of liquid solutions is defined as the pressure exerted by the vapours on the liquid solvent when kept in equilibrium and a certain temperature.
Question 2
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A mixture in which the mole ratio of $$H_{2}$$ and $$O_{2}$$ is $$2 : 1$$ is used to prepare water by the reaction,
$$2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)}$$
The total pressure in the container is $$0.8\ atm$$ at $$20^{\circ}C$$ before the reaction. The final pressure at $$120^{\circ}C$$ after reaction is: (assuming $$80$$% yield of water).

A
$$1.787\ atm$$

B
$$0.878\ atm$$

C
$$0.787\ atm$$

D
$$1.878\ atm$$

Solution

$$\underset {\underset {Final\ mode\ (2a - 2a)}{Initial\ mole\ 2a}}{2H_{2}(g)} + \underset {\underset {(a - x)}{a}}{O_{2(g)}}\rightarrow \underset {\underset {2x}{0}}{2H_{2}O_{(g)}}$$
Given $$2x = 2a \times \dfrac {80}{100} = 1.6a$$
$$\therefore x = 0.8a$$
Thus, after the reaction $$H_{2}\ left = 2a - 1.6a = 0.4a\ mole$$
$$O_{2}\ left = 0.2a\ mole$$
$$H_{2}O\ formed = 1.6a\ mole$$
$$\therefore$$ Total mole at $$120^{\circ}C$$ in gaseous phase
$$= 0.4a + 0.2a + 1.6a = 2.2a$$
Now, given at initial conditions $$P = 0.8\ atm, T = 293\ K$$
$$P\times V = nRT$$
$$0.8\ V = 3a \times R\times 293$$
$$\therefore V = \dfrac {3a\times R\times 293}{0.8}$$
The volume of container remains constant.
After the reaction using, $$P\times V = nRT$$
$$\therefore P\times \dfrac {3a\times R\times 293}{0.8} = 2.2a \times R\times 393$$
$$P = \dfrac {393\times 0.8\times 2.2}{3\times 293} = 0.787\ atm$$.
Question 3
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The certain volume of a gas exerts on its walls some pressure at a particular temperature. It has been found that by reducing the volume of the gas to half of its original value the pressure becomes twice that of the initial value at a constant temperature. This happens because:

A
mass of the gas increases with pressure

B
speed of the gas molecules decreases

C
more number of gas molecules strikes the surface per second

D
gas molecules attract each other

Solution

Certain volume of a gas exerts on its' walls some pressure at a particular temperature. It has been find that by reducing the volume of the gas to half of its' original value the pressure becomes twice that of the initial value at a constant temperature. This happens because- more number of gas molecules strikes the surface per second due to the reduce in volume at constant temperature.
Question 4
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The main reason for deviation of gases from ideal behaviour is few assumptions of kinetic theory. These are
(i) there is no force of attraction between the molecules of a gas
(ii) volume of the molecules of a gas is negligibly small in comparison to the volume of the gas
(iii) Particles of a gas are always in constant random motion.

A
(i) and (ii)

B
(ii) and (iii)

C
(i), (ii) and (iii)

D
(iii) only

Solution

The main reasons for deviation of gases from ideal behavior is few assumptions of Kinetic theory, they are-
$$\Longrightarrow$$ The kinetic theory of gases assume that gas molecules are point particles without any volume and any force acting between them; so that their collisions could be treated as perfectly elastic.
$$\Longrightarrow$$ But in reality, neither can one assure gas molecules to be with nil volume nor free from a mild attractive force.
Question 5
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When a non volatile solute is added to a pure solvent, the :

A
vapour pressure of the solution becomes lower then that of the pure solvent

B
rate of evaporation of the pure solvent is reduced

C
solute does not effect the rate of condensation

D
rate of the evaporation of the solution is equal to the rate of condensation of the solution at a lower vapour pressure than that in the case of the pure solvent

Solution

If we add a non volatile solute to solvent such
as water, we decease the tendency for water
molecules to evaporate into the gase phase.
in essence, the solute particles obstract the
evaporation of water, as a result, fcream
molecules change from the liquid to the gas
phase, thus reducing the vapour pressure.
$$ \boxed {hence\, Ans\, - \,'A' } $$
Question 6
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If $$P_o$$ and $$P_s$$ are the vapour pressure of solvent and its solution respectively. $$x_1$$ and $$x_2$$ are the mole fraction of solvent and solute respectively then:

A
$$P_s = \dfrac{P_o}{x_2}$$

B
$$P_o - P_s = P_ox_2$$

C
$$P_s = P_ox_2$$

D
$$\dfrac{(P_o-P_s)}{P_s} = \dfrac{x_1}{(x_1+x_2)}$$

Solution
Question 7
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5 litre of $$N_2$$ under a pressure of 2 atm, 2 litre of $$O_2$$ at 5.5 atm and 3 litre of $$CO_2$$ at 5 atm are mixed. The resultant volume of the mixture is 15 litre. Calculate the total pressure of the mixture and partial pressure of each constituent.

A
2.06 atm

B
1.75 atm

C
3.06 atm

D
4.50 atm

Solution

From the equation $$Pv = n RT$$ for the two gases. We can write
$$nN_2 = \dfrac{5 \times 2} {RT}$$

$$nO_2 = \dfrac{2 \times 5.5} {RT}$$

$$nCO_2 = \dfrac{3\times 5}{RT}$$
When introduced in $$1 L$$ vessel, then

$$P\times15L = (nN_2 + nO_2+nCO_2) RT$$
Putting the values, we get
$$P = 2.06 bar$$
Hence, the total pressure of the gaseous mixture in the vessel is $$1.8$$ bar
Question 8
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A 5 L vessel contains 1.4 g of nitrogen gas $$N_2$$. When heated to 1800 K, 30% of molecules are dissociated into atoms. Calculate the pressure of the gas at 1800 K. [Mol. mass of $$N_2 = 28 mol^{-1}$$, R = 0.0821 L atm $$K^{-1} mol^{-1}$$

A
2.45 atm

B
3.69 atm

C
1.92 atm

D
None of these

Solution

Since some of the nitrogen gas is dissociated into atoms the vessel will now contains a mixture of molecular nitrogen and atomic nitrogen, Let partial pressure as $$P_{1}$$ & $$P_{2}$$
mass of molecular nitrogen $$m_{1}=1.4\times \dfrac{70}{100}$$
$$=0.98\ g$$
mass of atomic nitrogen $$=m_{2}=1.4-0.98=0.42\ g$$
acc to Dalton law,
total pressure $$=p_{1}+p_{2}$$
$$p=p_{1}+p_{2}=\dfrac{n_{1}RT}{V}+\dfrac{n_{2}RT}{V}=\left(\dfrac{0.98}{28}\times\dfrac{0.0821\times 1800}{5}\right)+\left(\dfrac{0.42}{14}\times \dfrac{0.0821\times 1800}{5}\right)$$
$$\boxed{P=1.0344+0.887=1.9214\ atm}$$
Question 9
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Directions For Questions

A small spherical ball of radius $$r$$ is released from its completely submerged position (as shown in the figure) in a liquid whose density varies with height $$h$$ (measured from the bottom) as $${\rho}_{L}={\rho}_{0}[4-(3h/{h}_{0})]$$. The density of the ball is $$(5/2){\rho}_{0}$$ The height of the vessel is $${h}_{0}=12/{ \pi }^{ 2 }$$. Consider $$r << {h}_{0}$$ and $$g= 10\ m/{s}^{2}$$.

...view full instructions

The motion of the ball in the vessel is

A
Oscillatory but not simple harmonic motion

B
Simple harmonic motion with time period $$2\ s$$

C
Simple harmonic motion with time period $$1\ s$$

D
The motion is not oscillatory

Solution
Question 10
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Directions For Questions

The vapour pressure of two pure liquids $$A$$ and $$B$$ which form an ideal solution are $$500$$ and $$800$$ torr respectively at $$300K$$. A liquid solution of $$A$$ and $$B$$ for which the mole fraction of $$A$$ is $$0.60$$ is contained in a cylinder closed by a piston on which the pressure can be varied. The solution is slowly vaporized at $$300K$$ by decreasing the applied pressure.

...view full instructions

What is the composition of last deoplet of liquid remaining in equilibrium with vapour?

A
$$x_A = 0.6; x_B = 0.4$$

B
$$x_A = 0.5; x_B = 0.5$$

C
$$x_A = 0.7; x_B = 0.3$$

D
$$x_A = 0.3; x_B = 0.7$$

Solution

$$P_{A}=500$$ $$\quad P_{B}=800$$
$$x_{A}=0.6$$
$$x_{B}=1-x_{A}=0.4$$
To Find $$X_{A}$$ and $$X_{B}$$
where $$X_{A}=$$ Last drop of liquid remaining in Equilibrium in A.
$$X_{B}=$$ Last drop of liquid remaining in Equilibrium $B$
We Know that
Total pressure $$(P)=\quad P_{A}^{0} x_{A}+P_{B}^{0} x_{B}$$
$$=500 \times(0.6)+800 \times(0.4)=620$$
Now partial vapour pressure of $$A$$
$$\begin{aligned} y_{A}=\dfrac{P_{A}}{P} &=\dfrac{P_{A X A}^{0} }{P} \\ &=\dfrac{500 \times 0.6}{620}=0.48 \end{aligned}$$
Mole fraction of $$A=\dfrac{P_{A} x_{A}}{P_{A} x_{A}+P_{B}^{\circ}\left(1-x_{A}\right)}$$
$$0 .6=\dfrac{500 x_{A}}{500 x_{A}+800\left(1-x_{A}\right)}$$
we get
$$\begin{array}{rl}x_{A}=0.7 & x_{B}=1-x_{A} \\ & =0.3\end{array}$$

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States of Matter Test - 60

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States of Matter Test - 60

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Question 1

Vapour pressure in all the three beakers is same

Vapour pressure of beaker B is highest

Vapour pressure of beaker C is highest

Vapour pressure of beaker B is lower than that of C and vapour pressure of beaker C is lower than that of A

Solution

Question 2

$$1.787\ atm$$

$$0.878\ atm$$

$$0.787\ atm$$

$$1.878\ atm$$

Solution

Question 3

mass of the gas increases with pressure

speed of the gas molecules decreases

more number of gas molecules strikes the surface per second

gas molecules attract each other

Solution

Question 4

(i) and (ii)

(ii) and (iii)

(i), (ii) and (iii)

(iii) only

Solution

Question 5

vapour pressure of the solution becomes lower then that of the pure solvent

rate of evaporation of the pure solvent is reduced

solute does not effect the rate of condensation

rate of the evaporation of the solution is equal to the rate of condensation of the solution at a lower vapour pressure than that in the case of the pure solvent

Solution

Question 6

$$P_s = \dfrac{P_o}{x_2}$$

$$P_o - P_s = P_ox_2$$

$$P_s = P_ox_2$$

$$\dfrac{(P_o-P_s)}{P_s} = \dfrac{x_1}{(x_1+x_2)}$$

Solution

Question 7

2.06 atm

1.75 atm

3.06 atm

4.50 atm

Solution

Question 8

2.45 atm

3.69 atm

1.92 atm

None of these

Solution

Question 9

Oscillatory but not simple harmonic motion

Simple harmonic motion with time period $$2\ s$$

Simple harmonic motion with time period $$1\ s$$

The motion is not oscillatory

Solution

Question 10

$$x_A = 0.6; x_B = 0.4$$

$$x_A = 0.5; x_B = 0.5$$

$$x_A = 0.7; x_B = 0.3$$

$$x_A = 0.3; x_B = 0.7$$

Solution

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