States of Matter Test

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States of Matter Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

To which of the following the Dalton's law of partial pressures is not applicable?

A
$$H_{2}$$ and $$He$$

B
$$NH_{3}$$ and $$HCl$$

C
$$N_{2}$$ and $$H_{2}$$

D
$$N_{2}$$ and $$O_{2}$$

Solution

Daltons' law of partial pressure is applicable to non-reacting gases, $$NH_{3}$$ and $$HCl$$ are reacting gases, so Dalton's law will not be applicable for them.
Question 2
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A $$10\ L$$ flask contains a gaseous mixture of $$CO$$ and $$CO_{2}$$ at a total pressure of $$2\ atm$$ and $$298\ K$$. If $$0.20\ mole$$ of $$CO$$ is present, find its partial pressure.

A
$$0.49\ atm$$

B
$$1.51\ atm$$

C
$$1\ atm$$

D
$$2\ atm$$

Solution

Here, $$V=10L$$, $$n=0.20mole$$, and $$T=298K$$
By Ideal gas equation
$$PV=nRT$$
$$\therefore P = \dfrac {nRT}{V} = \dfrac {0.20\times 0.0821\times 298}{10} = 0.49\ atm$$.
Question 3
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If $$4\ moles$$ of an ideal gas at $$300\ K$$ occupy volume of $$89.6\ L$$, then pressure of the gas will be :

A
$$2\ atm$$

B
$$1\ atm$$

C
$$1.099\ atm$$

D
$$2.910\ atm$$

Solution

By Ideal gas equation,
$$PV = nRT$$
$$P = \dfrac {4\times 0.0821\times 300}{89.6} = 1.099\ atm$$.
Question 4
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A gas occupies a volume of $$300\ cm^{3}$$ at $$27^{\circ}C$$ and $$620\ mm$$ pressure. The volume of gas at $$47^{\circ}C$$ and $$640\ mm$$ pressure is:

A
$$260\ cm^{3}$$

B
$$310\ cm^{3}$$

C
$$390\ cm^{3}$$

D
$$450\ cm^{3}$$

Solution

Here, $$V_1=300cm^3$$, $$T=27^{\circ}C=300K$$, $$P_1=620mm$$
$$T_2=47^{\circ}C=320K$$, $$P_2=640mm$$
From Ideal gas equation,
$$\dfrac {P_{1}V_{1}}{T_{1}} = \dfrac {P_{2}V_{2}}{T_{2}} \Rightarrow \dfrac {620\times 300}{300} = \dfrac {640\times V_{2}}{320}$$
$$V_{2} = \dfrac {620\times 300\times 320}{640\times 300} = 310\ cm^{3}$$.
Question 5
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A closed container contains equal number of moles of two gases $$X$$ and $$Y$$ at a total pressure of $$710\ mm$$ of $$Hg$$. If gas $$X$$ is removed from the mixture, the pressure will :

A
become double

B
become half

C
remain same

D
become one-fourth

Solution

Here in the closed container, the equal number of moles of two gases are taken.
Thus, when gas $$X$$ is removed, the number of moles of the gases in the container becomes half.
According to an Ideal gas equation,
$$PV = nRT$$.
$$\therefore P \propto n$$
Thus, if all the other conditions are kept constant, the pressure is directly proportional to the number of moles.
Thus, if the number of moles of the gas is halved, the pressure will also become half of the original pressure.
Thus, B is the correct answer.
Question 6
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What will be the volume of $$2.8\ g$$ of carbon monoxide at $$27^{\circ}C$$ and $$0.821$$ atmospheric pressure?

A
$$2.5\ L$$

B
$$4\ L$$

C
$$3.5\ L$$

D
$$3\ L$$

Solution

Given, $$m=2.8\ g$$, $$T=27^{\circ}C=300K$$, and $$P=0.821\ atm$$
$$n = \dfrac {m}{M} = \dfrac {2.8}{28} = 0.1$$
By Ideal gas equation;
$$PV=nRT$$
$$\therefore V = \dfrac {nRT}{P} = \dfrac {0.1\times 0.0821\times 300}{0.821} = 3\ L$$.
Question 7
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$$34.05\ mL$$ of phosphorus vapours weight $$0.0625\ g$$ at $$546^{\circ}C$$ and $$0.1\ bar$$ pressure. What is the molar mass of phosphorus?

A
$$124.77\ g\ mol^{-1}$$

B
$$1247.74\ g\ mol^{-1}$$

C
$$12.47\ g\ mol^{-1}$$

D
$$30\ g\ mol^{-1}$$

Solution

Given, $$V=34.05mL$$, $$m=0.0625g$$, $$T=546^{\circ}C=819K$$
and $$P=0.1\ bar$$
Thus, molar mass of the phosphorus can be given as;
$$M = \dfrac {mRT}{PV} = \dfrac {0.0625\times 0.083\times 819}{0.1\times (34.05/ 1000)}$$
$$= 1247/74\ g\ mol^{-1}$$.
Question 8
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What will be the pressure of the gas mixture of $$3.2\ g$$ methane and $$4.4\ g$$ carbon dioxide contained in at $$9\ dm^{3}$$ flask at $$27^{\circ}C$$?

A
$$0.82\ atm$$

B
$$8.314\times 10^{4}\ atm$$

C
$$1\ atm$$

D
$$1.8\ atm$$

Solution

From Ideal gas law-
$$P = \dfrac {n}{V}RT$$

$$n$$ for $$CH_{4} = \dfrac {3.2}{16} = 0.2;$$

$$ n$$ for $$CO_{2} = \dfrac {4.4}{44} = 0.1$$

$$n_{total} = 0.2 + 0.1 = 0.3$$

$$P = \dfrac {0.3\times 0.0821\times 300}{9} = 0.82\ atm$$.

Hence, option A is correct.
Question 9
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Density of a gas is found to be $$5.46\ g/ dm^{3}$$ at $$27^{\circ}C$$ and $$2\ bar$$ pressure. What will be its density at STP?

A
$$3.0\ g\ dm^{-3}$$

B
$$5.0\ g\ dm^{-3}$$

C
$$6.0\ g\ dm^{-3}$$

D
$$10.82\ g\ dm^{-3}$$

Solution

$$P = \dfrac {n}{V}RT = \dfrac {m}{V} \times \dfrac {RT}{M} = \dfrac {dRT}{M}$$

or $$d = \dfrac {PM}{RT}$$ or $$d\propto \dfrac {P}{T}$$

$$\dfrac {d_{1}}{d_{2}} = \dfrac {P_{1}}{P_{2}}\times \dfrac {T_{2}}{T_{1}}$$ or

$$d_{2} = \dfrac {d_{1}\times P_{2} \times T_{1}}{P_{1}\times T_{2}}$$

$$d_{2} = \dfrac {5.46\times 1\times 300}{2\times 273} = 3.0\ g \ dm^{-3}$$

Therefore, the correct option is A.
Question 10
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The volume occupied by $$88\ g$$ of $$CO_2$$ at $$30^{\circ}C$$ and $$1\ bar$$ pressure will be:

A
$$5.05\ L$$

B
$$49.8\ L$$

C
$$2\ L$$

D
$$55\ L$$

Solution

$$w = 88\ g, M = 44\ g\ mol^{-1}, T = 273 + 30 = 303\ K$$
$$P = 1\ bar, V = ?$$
$$PV = nRT$$ or $$V = \dfrac {nRT}{P} = \dfrac {w}{m}\dfrac {RT}{P}$$
$$V = \dfrac {88\times 0.0821\times 303}{44\times 1} = 49.8\ L$$.

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States of Matter Test - 28

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States of Matter Test - 28

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Question 1

$$H_{2}$$ and $$He$$

$$NH_{3}$$ and $$HCl$$

$$N_{2}$$ and $$H_{2}$$

$$N_{2}$$ and $$O_{2}$$

Solution

Question 2

$$0.49\ atm$$

$$1.51\ atm$$

$$1\ atm$$

$$2\ atm$$

Solution

Question 3

$$2\ atm$$

$$1\ atm$$

$$1.099\ atm$$

$$2.910\ atm$$

Solution

Question 4

$$260\ cm^{3}$$

$$310\ cm^{3}$$

$$390\ cm^{3}$$

$$450\ cm^{3}$$

Solution

Question 5

become double

become half

remain same

become one-fourth

Solution

Question 6

$$2.5\ L$$

$$4\ L$$

$$3.5\ L$$

$$3\ L$$

Solution

Question 7

$$124.77\ g\ mol^{-1}$$

$$1247.74\ g\ mol^{-1}$$

$$12.47\ g\ mol^{-1}$$

$$30\ g\ mol^{-1}$$

Solution

Question 8

$$0.82\ atm$$

$$8.314\times 10^{4}\ atm$$

$$1\ atm$$

$$1.8\ atm$$

Solution

Question 9

$$3.0\ g\ dm^{-3}$$

$$5.0\ g\ dm^{-3}$$

$$6.0\ g\ dm^{-3}$$

$$10.82\ g\ dm^{-3}$$

Solution

Question 10

$$5.05\ L$$

$$49.8\ L$$

$$2\ L$$

$$55\ L$$

Solution

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