States of Matter Test

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States of Matter Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

A $$1.00$$L vessel containing $$1.00$$ g $$H_2$$ gas at $$27^0$$C is connected to a $$2.00$$ L vessel containing $$88.0$$g $$CO_2$$ gas, at also $$27^0$$C. When the gases are completely mixed, total pressure is:

A
$$20.5$$ atm

B
$$4.1$$ atm

C
$$16.4$$ atm

D
$$730.7$$ atm

Solution

$$P_{Total}=P_A+P_B$$ $$T=27+273=300K$$; $$n_{H_2}=\cfrac {1}{2}=0.5$$ moles
$$P_A=P_{H_2}=\cfrac {(nH_2)RT}{V}$$
$$=\cfrac {0.5 \times 0.082\times 300}{3}$$
$$=4.1 atm$$
$$P_B=P_{CO_2}=\cfrac {n_{CO_2}RT}{V}$$ $$n_{CO_2}=\cfrac {88}{44}=2$$ moles
$$=\cfrac {2 \times 0.082 \times 300}{3}$$
$$= 16.4 atm$$
$$\therefore P_{Total}=4.1+16.4=20.5 atm$$
Question 2
1 / -0

A vessel contains 0.5 mole each of $${ SO }_{ 2 }{ ,H }_{ 2 }$$ and $$CH_{ 4 }$$. Its outlet was made open and closed after sometime. Thus, order of partial pressure will be:

A
$${ PSO }_{ 2 }>{ PCH }_{ 4 }>{ PH }_{ 2 }$$

B
$${PH}_{2}>{PCH}_{4}>{PSO}_{2}$$

C
$${PH}_{2}>{PSO}_{2}>{PCH}_{4}$$

D
$${PH}_{2} ={PSO}_{2} = {PCH}_{4}$$

Solution

Given gases are $${ SO }_{ 2 },{ CH }_{ 4 }$$ and Hydrogen.
Molecular weight of given gases are as follows.
$${ SO }_{ 2 }>{ CH }_{ 4 }>{ H }_{ 2 }$$
As the molecular weight is high, Partial pressure is also high.
So, $${ PSO }_{ 2 }>{ PCH }_{ 4 }>{ PH }_{ 2 }$$
Question 3
1 / -0

An ideal gas obeying kinetic gas equation :

A
can be liquefied if its temperature is more than critical temperature

B
can be liquefied at any value of any value T and P

C
can not be liquefied under any value of T and P

D
can be liquefied if its pressure is more than critical pressure

Solution

An ideal gas is meant to have no intermolecular force of attraction
That's why a real gas has to reach, lowest pressure and highest temperature enough to overcome the intermolecular force of attraction and behave like an ideal gas.
So ideal gas can't be liquefied at any T and P.
Question 4
1 / -0

The vapour pressure of pure $$A$$ is $$10$$ torr and at the same temperature when $$1\ g$$ of $$B$$ is dissolved in $$20\ gm$$ of $$A$$, its vapour pressure is reduced to $$9.0$$ torr. If the molecular mass of $$A$$ is $$200$$ amu, then the molecular mass of $$B$$ is:

A
$$100\ amu$$

B
$$90\ amu$$

C
$$75\ amu$$

D
$$120\ amu$$

Solution

According to Dalton's law of partial pressure,

$$p_{total}=\chi_Ap_A+\chi_Bp_B$$

Since B is a non-volatile substance, $$p_B=0$$

$$\Rightarrow p_{total}=p_A\times \dfrac{\dfrac{m_A}{M_A}}{\dfrac{m_A}M_A+\dfrac{m_B}{M_B}}$$

$$\Rightarrow 9=10\times \dfrac{\dfrac{20}{200}}{\dfrac{20}{200}+\dfrac{1}{M_B}}$$

$$\Rightarrow M_B=90\ amu$$

Hence, option $$B$$ is the correct answer.
Question 5
1 / -0

A balloon filled with methane $$CH_4$$ is pricked with a sharp point and quickly plunged into a tank of hydrogen at the same pressure. After sometime the balloon will have?

A
Enlarged

B
Collapsed

C
Remained unchanged in size

D
Ethylene $$(C_2H_4)$$ inside it

Solution

Given: initially balloon has methane.
Balloon having methane is led to diffuse methane by pricking.
Again the same balloon suddenly plunged to a tank to effuse hydrogen to it.

According to graham's law of effusion/diffusion.
$$\dfrac { { r }_{ 1 } }{ { r }_{ 2 } } =\sqrt { \dfrac { { M }_{ 1 } }{ { M }_{ 2 } } } $$
$$\dfrac { { r }_{ { CH }_{ 4 } } }{ { r }_{ { H }_{ 2 } } } =\sqrt { \dfrac { 2 }{ 16 } } $$

We know, $$M =$$ Molecular weight
$$r =$$ Rate of effusion/diffusion
$$\dfrac { { r }_{ { CH }_{ 4 } } }{ { r }_{ { H }_{ 2 } } } = { \dfrac { 1 }{ 2\sqrt { 2 } } } $$

$${ r }_{ { H }_{ 2 } }=2\sqrt { 2 } \times { r }_{ { CH }_{ 4 } }$$

Rate of hydrogen effusion is $$2\sqrt { 2 } $$ times the rate of methane diffusion.
Therefore, balloon enlarges as hydrogen moves in faster.
Question 6
1 / -0

At a pressure of 760 torr and temperature of 273.15K, the indicated volume of which system is not consistent with the observation?

A
14 g of $$N_2$$ + 16 g of $$O_2$$; Volume = 22.4 L

B
4 g of He + 44 g of $$CO_2$$; Volume = 44.8 L

C
7 g of $$N_2$$ + 36 g of $$O_3$$; Volume = 22.4 L

D
17 g of $$NH_3$$ + 36.5 g of $$HCl$$, Volume = 44.8 L

Solution

Given,
Pressure=1atm
temperature=$$73.15K$$
Now,For volume calculation,we know
1 mole of any gaseous mixture=$$22.4L$$
n mole of any gaseous mixture$$=n\times 22.4L$$
Volume of mixture$$=n\times 22.4$$
Now,
a)14g of $${N}_{2}$$+16g of $${O}_{2}$$
$$=\cfrac{14}{28}+\cfrac{16}{32}=\cfrac{1}{2}+\cfrac{1}{2}=1$$ mole
Volume = $$22.4L$$
b) 4g of $$He+44g$$ of $${CO}_{2}$$
$$=\cfrac{4}{4}+\cfrac{44}{44}=1+1=2$$mole
Volume$$=2\times 22.4=44.8L$$
c)7g of $${N}_{2}$$+36g of $${O}_{3}$$
$$=\cfrac{7}{28}+\cfrac{36}{48}=\cfrac{1}{4}+\cfrac{3}{4}=1$$mole
Volume$$=22.4L$$
d)17g of $${NH}_{3}$$+36.5g of HCl
$$=\cfrac{17}{17}+\cfrac{36.5}{36.5}=2$$mole
Volume of mixture$$=44.8L$$
Because $${ NH }_{ 3 }(g)+HCl(g)\longrightarrow { NH }_{ 4 }Cl(s)$$
Volume of solid<<volume of gas
Therefore, Volume of mixture is $$<<44.8L$$
Question 7
1 / -0

The volume of helium gas is 44.8 L at:

A
100$$^oC$$ and 1 atm

B
0 $$^o C$$ and 1 atm

C
0 $$^o \ C$$ and 0.5 atm

D
100 $$^o \ C$$ and 0.5 atm

Solution

The conditions at STP are:

$$P=1\ atm$$
$$T=273.15\ K$$

We know that the volume occupied by any gas at STP is $$22.4\ L$$

Hence using the relation $$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$,

$$\dfrac{1\times22.4}{273.15}=\dfrac{P_2\times44.8}{T_2}$$

$$\dfrac{T_2}{P_2}=546.3$$

Since this condition is satisfied by a pressure of 0.5 atm and a temperature of 273 K.
Hence option $$C$$ is the correct answer
Question 8
1 / -0

How much should the pressure be increased in order to decrease the volume of a gas by $$15\%$$ at a constant temperature?

A
$$15.6\%$$

B
$$17.6\%$$

C
$$6.7\%$$

D
$$8.9\%$$

Solution

Given,volume is decreased by 15%
$${V}_{1}=V$$
$${V}_{2}=V-15V/100$$
$$\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =1.176$$ ---------1
Now,we know
$$PV=nRT$$
at T constant,R constant and n constant
PV=K
$$P\propto \dfrac { 1 }{ V } $$
$$\dfrac { { P }_{ 2 } }{ { P }_{ 1 } } =\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =1.176$$
Now % of pressure increased:$$(\dfrac { { P }_{ 2 } }{ { P }_{ 1 } } -1)\times 100$$
$$=(1.176-1)\times100$$
$$=17.6\%$$
Question 9
1 / -0

A gaseous mixture contain 1 gm of $${ H }_{ 2},$$ 4 gm of $$He$$ 7gm of $${ N }_{ 2 } $$ and 8gm of $${ O }_{ 2 }$$. The gas having the highest partial pressure is:

A
$${ H }_{ 2 }$$

B
$${ O }_{ 2 }$$

C
$$He$$

D
$${ N }_{ 2 }$$

Solution

Given,
1gm of hydrogen, 4gm of helium,7gm of nitrogen,8gm of oxygen
Now,
Given mole of hydrogen$$\cfrac{1}{2}$$mole
mole of helium$$=\cfrac{4}{4}$$=1mole
mole of nitrogen$$=\cfrac{7}{28}=\cfrac{1}{4}$$ mole
mole of oxygen$$=\cfrac{8}{32}=\cfrac{1}{4}$$mole
We know according to ideal gas equation
$$P\propto n$$ , P=pressure, n=no of moles
Therefore,the gas having highest no of moles will have height partial pressure.
hence helium will have highest partial pressure.
Question 10
1 / -0

An aquatic species need at least 4mg/L of $$O_{2}$$ for their survival. $$O_{2}$$ in water at 273K and 1 atm pressure is $$2.21 \times 10^{-3}$$ mol/L. The partial pressure of $$O_{2}$$ above water(in an atmosphere at 273K) needed for the survival of species is:

A
0.56 atm

B
0.056 atm

C
5.525 atm

D
5.525 $$\times $$10$$^{-3}$$ atm

Solution

Given, required solubility of oxygen gas for the survival of marine species=4mg/L
We know 1 mole of oxygen $$= 32g$$ of oxygen
0.004g of $${O}_{2}=125 \times {10}^{-6}mol/L$$
Now we know initial pressure above water = 1 atm
Also given,
Initially solubility of $${O}_{2}$$ gas in water at 1 atm $$=2.21 \times {10}^{-3}mol/L$$
Now,we know according to Henry's law of partial pressure.
Solubility of gas is directly proportional to partial pressure of that gas over liquid.
$$S\propto P$$

$$\dfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } $$

$$\dfrac { 2.21\times { 10 }^{ -3 } }{ 125\times { 10 }^{ -6 } } =\dfrac { 1 }{ { P }_{ 2 } } $$

$${P}_{2}=0.056atm$$

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States of Matter Test - 39

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States of Matter Test - 39

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Question 1

$$20.5$$ atm

$$4.1$$ atm

$$16.4$$ atm

$$730.7$$ atm

Solution

Question 2

$${ PSO }_{ 2 }>{ PCH }_{ 4 }>{ PH }_{ 2 }$$

$${PH}_{2}>{PCH}_{4}>{PSO}_{2}$$

$${PH}_{2}>{PSO}_{2}>{PCH}_{4}$$

$${PH}_{2} ={PSO}_{2} = {PCH}_{4}$$

Solution

Question 3

can be liquefied if its temperature is more than critical temperature

can be liquefied at any value of any value T and P

can not be liquefied under any value of T and P

can be liquefied if its pressure is more than critical pressure

Solution

Question 4

$$100\ amu$$

$$90\ amu$$

$$75\ amu$$

$$120\ amu$$

Solution

Question 5

Enlarged

Collapsed

Remained unchanged in size

Ethylene $$(C_2H_4)$$ inside it

Solution

Question 6

14 g of $$N_2$$ + 16 g of $$O_2$$; Volume = 22.4 L

4 g of He + 44 g of $$CO_2$$; Volume = 44.8 L

7 g of $$N_2$$ + 36 g of $$O_3$$; Volume = 22.4 L

17 g of $$NH_3$$ + 36.5 g of $$HCl$$, Volume = 44.8 L

Solution

Question 7

100$$^oC$$ and 1 atm

0 $$^o C$$ and 1 atm

0 $$^o \ C$$ and 0.5 atm

100 $$^o \ C$$ and 0.5 atm

Solution

Question 8

$$15.6\%$$

$$17.6\%$$

$$6.7\%$$

$$8.9\%$$

Solution

Question 9

$${ H }_{ 2 }$$

$${ O }_{ 2 }$$

$$He$$

$${ N }_{ 2 }$$

Solution

Question 10

0.56 atm

0.056 atm

5.525 atm

5.525 $$\times $$10$$^{-3}$$ atm

Solution

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