States of Matter Test

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States of Matter Test - 51

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Question 1
1 / -0

1 mole of chlorine and 1 mole of hydrogen were taken in a 10 L evacuated flask along with a little charcoal. The flask was then irradiated with light until the reaction was complete. Subsequently, 5L of water was introduced into the flask and the flask was cooled to $$27^{o}C$$. The pressure exerted by the system is approximately equal to:

[Aqueous tension at $$27^{o}C = 26 \ mm$$]

A
98.5 mm

B
26 mm

C
52 mm

D
76 mm

Solution

Since, $$HCl$$ is highly soluble in water, no gas is present in the flask. The pressure exerted is by water vapour which is equal to aqueous tension i.e., 26 mm at $$27^oC$$.
Question 2
1 / -0

According to the kinetic theory of gases, in an ideal gas, between two successive collisions, a gas molecule travels:

A
in a wavy path

B
in a straight line path

C
with an accelerated velocity

D
in a circular path

Solution

According to the postulates of kinetic theory of gases, the molecules of a gas move in a straight line between two successive collisions. They change their path only when they suffer collisions with the other molecules or with the walls of the container.
Question 3
1 / -0

An open vessel at $$27^oC$$ is heated until the three-fourths mass of the air in it has been expelled. Neglecting the expansion of the vessel, find the temperature to which the vessel has been heated.

A
$$T_2=723^oC$$

B
$$T_2=958^oC$$

C
$$T_2=927^oC$$

D
None of these

Solution

No. of moles of air$$=n$$.

$$\cfrac{3}{4}$$ of original moles has expelled.

Remaining moles$$=n(1-\cfrac{3}{4})$$
$$=\cfrac{n}{4}$$.

At constant pressure & volume,

$$n_{1}T_{1}=n_{2}T_{2}$$.

$$\Rightarrow (n)(300)=(\cfrac{n}{4})(T_{2})$$

$$\Rightarrow T_{2}=(300)(4)$$
$$=1200\,K$$

$$\Rightarrow T_{2}=927^{\circ}C$$.

$$\therefore$$ Temperature of vessel is $$927^{\circ}C$$.

Hence, (C) is the correct option.
Question 4
1 / -0

Calculate the volume of $$CO_{2}$$ evolved by the combustion of $$50$$ ml of a mixture containing $$40$$ per $$C_{2}H_{4}$$ and $$60$$ per $$CH_{4}$$ (by volume).

A
$$70$$ ml

B
$$75$$ ml

C
$$80$$ ml

D
$$82$$ ml

Solution

The volume of $$C_2H_4$$ in $$50$$ ml of sample is $$50 ml \times \dfrac {40} {100} =20 ml$$.

The volume of $$CH_4$$ in $$50$$ ml of sample is $$50 ml \times \dfrac {60} {100} =30 ml$$.

$$\underset {30 ml}{CH_4} \rightarrow \underset {30 ml}{CO_2} + \underset {60 ml} {2H_2O} $$

$$\underset {20 ml}{C_2H_4} \rightarrow \underset {40}{2CO_2} + \underset {40 ml} {2H_2O} $$
The volume of carbon dioxide formed is $$30 ml + 40 ml=70 ml$$.
Question 5
1 / -0

$$45.4$$ L of dinitrogen reacted with $$22.7$$ L of dioxygen and $$45.4$$ L of nitrous oxide was formed. The reaction is given below:
$$\displaystyle 2N_{2}\left ( g \right )+O_{2}\left ( g \right )\rightarrow 2N_{2}O\left ( g \right )$$
Which law is being obeyed in this experiment?

A
Gay Lussac's law of gaseous volumes

B
Henry's Law

C
Raoult's Law

D
None

Solution

The ratio of the volumes of dinitrogen, oxygen and nitrous oxide is $$45.4$$ : $$22.7$$ : $$45.4$$ . This is in the simple whole number ratio $$2:1:2$$.
This satisfies the Gay Lussac's law of combining volume of gases.
Question 6
1 / -0

Two identical vessels are filled with 44 g of hydrogen and 44 g of carbon dioxide at the same temperature. If pressure of $$CO_2$$ is 2 atm. Find the pressure of hydrogen.

A
$$P_{H_2}=56\ atm$$

B
$$P_{H_2}=44\ atm$$

C
$$P_{H_2}=32\ atm$$

D
None of these

Solution

Two identical vessels,

$$44\,g$$ of $$CO_{2}=1\,mol$$ and exerts $$2\,atm$$ pressure.

Now,
$$44\,g$$ of $$H_{2}=\cfrac{44}{2}=22\,mol$$ and hence, will exert $$={22}\times{2}\,atm$$ pressure.

$$\therefore$$ Pressure of $$H_{2}=44\,atm$$.
Question 7
1 / -0

A 600 $$ml$$ vessel containing oxygen at 800 $$mm$$ and a 400 $$ml$$ vessel containing nitrogen at 600 $$mm$$ at the same temperature are joined to each other. The final pressure of the mixture is:

A
1400 $$mm$$

B
1000 $$mm$$

C
720 $$mm$$

D
700 $$mm$$

Solution

For an ideal gas equation;
$$PV = nRT$$

$$\therefore$$ For $$O_{2}$$ , no. of moles = $$\dfrac{PV}{RT}$$ = $$\dfrac{0.6\times 800}{RT} = \dfrac{480}{RT}$$

For $$N_{2}$$, no. of moles = $$\dfrac{0.4\times 600}{RT} = \dfrac{240}{RT}$$

Total no. of moles = $$\dfrac{480}{RT} + \dfrac{240}{RT} = \dfrac{720}{RT}$$ = n

$$\therefore$$ Pressure of final mixture = $$\dfrac{nRT}{Total volume} = \dfrac{720}{RT}\times \dfrac{RT}{1}$$ = 720 $$mm$$
Question 8
1 / -0

A balloon of diameter $$21\:meter$$ weight $$100 kg$$. Calculate its pay-load, if it is filled with $$He$$ at $$1.0\: atm$$ and $$27 ^\circ C$$. Density of air is $$1.2\: kg m^{-3}$$. (Given: $$R = 0.082\: L\: atm\: K^{-1} mol^{-1}$$)

A
4952.42 kg

B
4932.42 kg

C
493.242 kg

D
None of these

Solution

Weight of balloon$$=100\:kg=10 \times 10^{4}\:g$$.

Volume of balloon$$= \displaystyle \frac{4}{3} \pi r^{3}$$.

$$= \displaystyle \frac{4}{3} \times \displaystyle \frac{22}{7} \times ( \displaystyle \frac{21}{2} )^{3}$$

$$=4851 m^{3}=4851 \times 10^{3} \:L$$

Weight of gas $$(He)$$ in balloon$$= \displaystyle \frac{PVM}{RT}$$

$$= \displaystyle \frac{1 \times 4851 \times 10^{3} \times 4}{0.082 \times 300} =78.878 \times 10^{4}$$

$$\therefore$$ Total weight of gas and balloon
$$=78.878 \times 10^{4}+10 \times 10^{4}$$

$$=88.878 \times 10^{4}g$$ or $$888.78\:kg$$

Weight of air displaced$$=1.2 \times 4851$$
$$=5821.2\:kg$$

$$\therefore$$ Pay load=wt. of air displaced-wt. of (balloon+gas)
$$\therefore$$ Pay load$$=5821.2-888.78=4932.42\:kg$$.
Question 9
1 / -0

Correct expression for density of an ideal gas mixture of two gases 1 and 2, where $$m_{1}$$ and $$m_{2}$$ are masses and $$n_{1}$$ and $$n_{2}$$ are moles and $$M_{1}$$ and $$M_{2}$$ are molar masses.

A
$$d=\displaystyle \frac{(m_{1}+m_{2})}{(M_{1}+M_{2})}$$

B
$$d=\displaystyle \frac{(m_{1}+m_{2})}{(n_{1}+n_{2})} \displaystyle \frac{P}{RT}$$

C
$$d=\displaystyle \frac{(n_{1}+n_{2})}{(m_{1}+m_{2})} \times \displaystyle \frac{P}{RT}$$

D
None of these

Solution

By ideal gas equation $$PV=nRT; d = (P/RT) \times 1/M$$
So for a mixture of two gases,
$$d=\displaystyle \frac{(m_{1}+m_{2})}{(n_{1}+n_{2})} \displaystyle \frac{P}{RT}$$
Question 10
1 / -0

For any gas, the mean free path at a particular pressure :

A
Independent of temperature

B
Decreases with rise in temperature

C
Increases with rise in temperature

D
Directly proportional to $$T^{2}$$

Solution

As the temperature increases, the increase in the distance traveled per unit time is greater than the increase in the number of collision made by a single molecule per unit time.

Hence, the mean free path increases with an increase in temperature.

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 51

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States of Matter Test - 51

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SHARING IS CARING

Question 1

98.5 mm

26 mm

52 mm

76 mm

Solution

Question 2

in a wavy path

in a straight line path

with an accelerated velocity

in a circular path

Solution

Question 3

$$T_2=723^oC$$

$$T_2=958^oC$$

$$T_2=927^oC$$

None of these

Solution

Question 4

$$70$$ ml

$$75$$ ml

$$80$$ ml

$$82$$ ml

Solution

Question 5

Gay Lussac's law of gaseous volumes

Henry's Law

Raoult's Law

None

Solution

Question 6

$$P_{H_2}=56\ atm$$

$$P_{H_2}=44\ atm$$

$$P_{H_2}=32\ atm$$

None of these

Solution

Question 7

1400 $$mm$$

1000 $$mm$$

720 $$mm$$

700 $$mm$$

Solution

Question 8

4952.42 kg

4932.42 kg

493.242 kg

None of these

Solution

Question 9

$$d=\displaystyle \frac{(m_{1}+m_{2})}{(M_{1}+M_{2})}$$

$$d=\displaystyle \frac{(m_{1}+m_{2})}{(n_{1}+n_{2})} \displaystyle \frac{P}{RT}$$

$$d=\displaystyle \frac{(n_{1}+n_{2})}{(m_{1}+m_{2})} \times \displaystyle \frac{P}{RT}$$

None of these

Solution

Question 10

Independent of temperature

Decreases with rise in temperature

Increases with rise in temperature

Directly proportional to $$T^{2}$$

Solution

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