States of Matter Test

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States of Matter Test - 34

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Question 1
1 / -0

Air contains $$79\% \ N_2$$ and $$21\% O_2$$ by volume. If the pressure is $$750 \ mm$$ of $$Hg$$, the partial pressure of oxygen $$(O_2)$$ is:

A
$$157.5\ mmHg$$

B
$$175.5\ mmHg$$

C
$$315.0\ mmHg$$

D
$$257.5\ mmHg$$

Solution

Mole Fraction of Oxygen = 0.21
Barometric Pressure = 750 mg
Partial Pressure of oxygen = Mole Fraction x Barometric Pressure
Partial Presuure of Oxygen $$= 0.21 \times 750$$
$$= 157.5\ mmHg$$
Question 2
1 / -0

A balloon weighting $$50 Kg$$ is filled with $$ 685 kg$$ of helium at $$1 atm$$ pressure and $${ 25 }^{ o }C$$. What will be its pay load if it displaced $$5108 kg$$ of air?

A
$$4373 kg$$

B
$$4423 kg$$

C
$$5793 kg$$

D
None of these

Solution

The payload can be explained as the difference between the mass of the air displaced and that of the mass of balloon whereas mass of balloon is equal to the sum of the mass of balloon skin and mass of Helium filled in it.
Given that:-
Mass of air displaced $$= 5108 \; Kg$$
mass of balloon skin $$= 50 \; Kg$$
Mass of helium filled in it $$= 685 \; Kg$$
$$\therefore \text{ Payload} = 5108 - \left( 50 + 685 \right) = 5108 - 735 = 4373 \; Kg$$
Question 3
1 / -0

The measured density at $$NTP$$ of a gaseous sample of a compound was found to be $$1.78\ g/L$$. What is the weight of $$1\ mole$$ of the gaseous sample?

A
$$45.6\ g$$

B
$$39.9\ g$$

C
$$28.25\ g$$

D
None of these

Solution

$$1.78g$$ of gas$$\rightarrow$$ 1lt at $$NTP\\$$
$$xgram\rightarrow 39.9g\\$$
$$x=1.789\times 39.9\\$$
$$x=71g$$
so it is iron sulphate,which is having $$70g$$ of molecular weight.
conclusion: hence the answer is option(D) None of these.
Question 4
1 / -0

$$2.8 g$$ of a gas at $$1atm$$ and $$273K$$ occupies a volume of $$2.24$$ litres. The gas can not be:

A
$${ O }_{ 2 }$$

B
$$CO$$

C
$${ N }_{ 2 }$$

D
$$C_{ 2 }{ H }_{ 4 }$$

Solution

Let the molecular weight of gas be M.
Weight of gas $$= 2.8 g$$
no. of moles of gas, $$\left( n \right) = \cfrac{\text{wt.}}{\text{mol. wt.}} = \cfrac{2.8}{M}$$
Pressure, $$\left( P \right) = 1 atm$$
Volume, $$\left( V \right) = 2.24 L$$
Temperature, $$\left( T \right) = 273 K$$
$$R = 0.082 \; L \; atm \; {mol}^{-1} \; {K}^{-1}$$
From ideal gas equation,
$$PV = nRT$$
$$\Rightarrow \; 1 \times 2.24 = \cfrac{2.8}{M} \times 0.082 \times 273$$
$$\Rightarrow \; M = \cfrac{62.68}{2.24}$$
$$\Rightarrow \; M = 27.98 g \approx 28 g$$
Hence the molecular mass of gas must be 28 g.
Since molecular mass of $${O}_{2}$$ is 32 g, hence the gas cannot be $${O}_{2}$$.
Question 5
1 / -0

A $$3.00\ L$$ oxygen gas is collected over water at $${27}^{o}C$$ when the barometric pressure is $$787\ Torr$$. If vapour pressure of water is $$27\ Torr$$ at $${27}^{o}C$$, moles of $$O_{2}$$ gas will be?

A
$$0.122$$

B
$$0.126$$

C
$$0.004$$

D
$$0.152$$

Solution

Given Volume (V)$$=3.00$$ L
Temperature(T)$$=27^{0}C=300 K$$
Barometric pressure$$=787$$ torr
Vapour pressure of water$$=27$$ torr
total pressure$$=787-27=760\quad{torr}=1\quad{atm}$$
Applying in ideal gas equation
PV=nRT
$$n=\dfrac{PV}{RT}$$
$$n=\dfrac{{1}\times{3}}{{0.0821}\times{300}}=0.122$$
Question 6
1 / -0

A liquid A and B form an ideal solution. If vapour pressure of pure A and B are $$500N{ m }^{ -2 }$$ and $$200N{ m }^{ -2 }$$ respectively, the vapour pressure of a solution of A and B containing 0.2 mole fraction of A would be:

A
$$700N{ m }^{ -2 }$$

B
$$300N{ m }^{ -2 }$$

C
$$260N{ m }^{ -2 }$$

D
$$140N{ m }^{ -2 }$$

Solution

Vapour pressure of a solution containing A & B liquid is
$$P_T=P_A+P_B$$
$$P_A= X_A-P^o_A$$ $$X_A$$= mole fraction of A; $$P^o_A$$= Pure vapour pressure of A.
$$\Rightarrow P_A= 0.2 \times 500 Nm^{-2}$$ & $$P_B= (1-0.2)\times 200 Nm^{-2}$$
Thus $$P_T= 0.2 \times 500 Nm^{-2} + 0.8 \times 200 Nm^{-2}$$
$$\Rightarrow P_T= 260 Nm^{-2}$$
Thus total vapour pressure of solution is $$260 Nm^{-2}$$ .
Question 7
1 / -0

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $$290\ mm$$ at $$300\ K$$. The vapour pressure of propyl alcohol is $$200\ mm$$. If the mole fraction of ethyl alcohol is $$0.6$$, its vapour pressure (in $$mm$$) at the same temperature will be:

A
$$300$$

B
$$700$$

C
$$360$$

D
$$350$$

Solution

By Roult's law,
$${ P }_{ T }={ P }_{ A }^{ \circ }{ x }_{ A }+{ P }_{ B }^{ \circ }{ x }_{ B }$$

$$\therefore \quad 290mm={ P }_{ A }^{ \circ }\times 0.6+200\times \left( 1-0.6 \right) $$

$$={ P }_{ A }^{ \circ }\times 0.6+80$$

$$\therefore \quad { P }_{ A }^{ \circ }=350\ mm$$

Hence, the correct option is $$D$$
Question 8
1 / -0

Equal weight of methane and hydrogen are mixed in an empty container at $${ 25 }^{ 0 }C$$. The fraction of total pressure exerted by hydrogen is?

A
1/2

B
8/9

C
16/19

D
1/9

Solution

Molar ratio of $$CH_4$$ : $$H_2$$ = 16 : 2 = 8 : 1 (Mixed in equal quantities)
Pressure is directly proportional to number of moles.
Methane is 8/9 of total number of moles.
So,
Fraction of pressure by $$CH_4$$ = 8/9
Therefore B is the correct option.
Question 9
1 / -0

A baloon blown up with $$1 $$ mole of gas has a volume of $$480 mL$$ at $$5^oC.$$ the balloon is field to $$(7/8)th$$ of its maximum capacity suggest. The minimum temperature at which it will brust?

A
$$31.4^oC$$

B
$$44.7^o C$$

C
$$49.1^o C $$

D
$$37.6^o C$$

Solution

Given,
$$n_1=1$$ mole
$$R= 0.0821$$
$$T_1=5^oC= 278K$$
$$V_1=480ml= 0.48L$$
$$P_1= \cfrac {n_1RT_1}{V_1}=\cfrac {1\times 0.0821 \times 278}{0.48}$$
$$\Rightarrow P_1= 47.55 atm$$
Now, as the given volume i.e. $$480ml$$ is $$7/8^{th}$$ of the total capacity.
Thus total volume will be
$$\Rightarrow 480ml \times \cfrac {8}{7}= 548.57 ml$$
The new temperature at which the balloon can burst will be at highest volume of balloon,
$$P_1V_2=n_1RT_2$$
$$\Rightarrow 47.55 atm \times 548.57 ml= 1\times 0.0821 \times T_2$$
$$\Rightarrow T_2= 317.7 K$$
$$\Rightarrow T_2= 44.7^oC$$
This is the minimum temperature at which balloon burst.
Question 10
1 / -0

The surface of a lake is at $$2^{o}C$$ and depth of the lake is $$20\ m$$. Find the temperature of the bottom of the lake?

A
$$552^{o}C$$

B
$$42^{o}C$$

C
$$4^{o}C$$

D
$$825^{o}C$$

Solution

It depends upon a number of factors like- how deep is the lake, how long has the temperature been at 2C, is there any movement in the water, are there any streams entering or leaving the lake .
For deep lakes with no heat sources and no movement, the temperature at the bottom will be 3.98C, because water at that temperature is at its most dense.

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 34

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States of Matter Test - 34

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SHARING IS CARING

Question 1

$$157.5\ mmHg$$

$$175.5\ mmHg$$

$$315.0\ mmHg$$

$$257.5\ mmHg$$

Solution

Question 2

$$4373 kg$$

$$4423 kg$$

$$5793 kg$$

None of these

Solution

Question 3

$$45.6\ g$$

$$39.9\ g$$

$$28.25\ g$$

None of these

Solution

Question 4

$${ O }_{ 2 }$$

$$CO$$

$${ N }_{ 2 }$$

$$C_{ 2 }{ H }_{ 4 }$$

Solution

Question 5

$$0.122$$

$$0.126$$

$$0.004$$

$$0.152$$

Solution

Question 6

$$700N{ m }^{ -2 }$$

$$300N{ m }^{ -2 }$$

$$260N{ m }^{ -2 }$$

$$140N{ m }^{ -2 }$$

Solution

Question 7

$$300$$

$$700$$

$$360$$

$$350$$

Solution

Question 8

1/2

8/9

16/19

1/9

Solution

Question 9

$$31.4^oC$$

$$44.7^o C$$

$$49.1^o C $$

$$37.6^o C$$

Solution

Question 10

$$552^{o}C$$

$$42^{o}C$$

$$4^{o}C$$

$$825^{o}C$$

Solution

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