States of Matter Test

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States of Matter Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following relationships is valid?

A
$$\dfrac {pV}{T}=\dfrac {2}{3}K$$

B
$$\dfrac {pV}{T}=\dfrac {3}{2}K$$

C
$$\dfrac {pV}{T}=2K$$

D
$$\dfrac {pV}{T}=3K$$

Solution

$$pV=\dfrac{1}{3}mN{C}_{2}$$ ($$mN=$$ molar mass)
For 1 mole of gas pV=RT, n=N(Av. constant)
$$RT =\dfrac { 1 }{ 3 }mN{C}_{2}$$
$$RT =\dfrac{2}{3}\times\dfrac{1}{2}mN{c}_{2}$$
$$=\dfrac{2}{3}$$K
So, $$pV=\dfrac{2}{3} $$K
$$C=$$ Speed of molecule
$$p=$$ Pressure
$$V=$$ Volume
$$R=$$ Gas constant
$$T=$$ Temperature
$$n=$$ No. of moles
$$m=$$ Molar mass
Question 2
1 / -0

Which of the following statement is not a postulate of kinetic molecular theory of gases?

A
$$K.E.$$ is dependent on temperature

B
$$K.E.$$ is dependent on pressure

C
The molecular collisions are perfectly elastic collisions

D
The pressure of the gas is due to collisions of gas molecules on the walls of the vessel

Solution

Pressure does not have any effect on the kinetic energy of the molecule. Gas pressure is due to the molecules colliding to the walls of the container. the kinetic energy of the molecule directly proportional to the temperature.
Question 3
1 / -0

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is $$1:4$$. The ratio of the number of their molecule is _______.

A
$$1:4$$

B
$$7:32$$

C
$$1:8$$

D
$$3:16$$

Solution

We know
Moles ratio $$=$$ number of molecules ratio

$$\dfrac { { n }_{ { O }_{ 2 } } }{ { n }_{ { N }_{ 2 } } } =\dfrac { \dfrac { m_{O_2} }{ { M }_{ { O }_{ 2 } } } }{ \dfrac { { m }_{ { N }_{ 2 } } }{ { M }_{ { N }_{ 2 } } } } $$
Where,
$${ m }_{ O_{ 2 } }=$$ given mass of $${ O }_{ 2 }$$
$${m }_{ { N }_{ 2 } }=$$ given mass of $${N}_{2}$$
$${M}_{O_{2}}=$$ Molecular mass of $${ O }_{ 2 }$$
$${M}_{N_{2}}=$$ molecular mass of $${N}_{2}$$
$${n }_{ { O }_{ 2 } }=$$ Number of moles of $${ O }_{ 2 }$$
$${n }_{ { N }_{ 2 } }=$$ Number of moles of $${N}_{2}$$

$$ =\left[ \dfrac { { m }_{ { O }_{ 2 } } }{ { m }_{ { N }_{ 2 } } } \right] \dfrac { 28 }{ 32 } =\dfrac { 1 }{ 4 } \times \dfrac { 28 }{ 32 } =\dfrac { 7 }{ 32 } $$

Therefore, the answer is 7:32
Question 4
1 / -0

For which of the following binary solution $${P}_{Total}={P}_{A}^{o}\times {X}_{A}+ {P}_{B}^{o}\times {X}_{B}$$ is followed?

A
Phenol+ Aniline

B
Benzene + Toluene

C
Acetone + Chloroform

D
water + $$HCl$$

Solution

The given expression is followed by ideal solutions where there are no attractive forces between the components.
Out of the given options, only option B (Benzene + Toluene) is an ideal one as in other solutions, there exists hydrogen bonding between the two components.
Question 5
1 / -0

Which gas does have density $$1,58$$ by air?

A
$$NO_2$$

B
$$N_2O$$

C
$$N_2O_3$$

D
$$NO$$

E
$$CO_2$$

Solution

$$N_2O$$ molecular weight is 44
Density of N$$_2$$O at STP =$$\dfrac{44}{22.4}$$
Air has 78% of $$N_2$$ . Its density =$$\dfrac{28}{22.4}$$
Relative density of $$N_2$$O with air is =$$\dfrac{Density \ of\ N_2O}{Density\ of\ N_2\ in \(air)}$$
=$$\dfrac{44/22.4}{28/22.4}=\dfrac{44}{28}$$=1.58
$$N_2O$$ has density 1.58 that of air.
Question 6
1 / -0

The mass of glucose that should be dissolved in $$100g$$ water in order to produce same lowering of vapour pressure as produced by dissolving $$1g$$ urea in $$50g$$ of water is?

A
$$1g$$

B
$$2g$$

C
$$6g$$

D
$$12g$$

Solution

$$\left( { H }_{ 2 }NCON{ H }_{ 2 } \right) \rightarrow M=60g/mol$$
$$m=$$ molar mass
$${ M }_{ { C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 } }=12\times 6+12\times 1+6\times 6=72+12+96+180 g/mol$$
Relative lowering in Vapour pressure$$=no.\quad of\quad moles\times \cfrac { molar\quad mass\quad of\quad solvent }{ mass\quad of\quad water } \\ =0.33\times \cfrac { 18 }{ 50 } $$
$$\cfrac { W }{ 180 } \times \cfrac { 18 }{ 100 } =\cfrac { 1 }{ 60 } \times \cfrac { 18 }{ 50 } \\ W=6g$$
Question 7
1 / -0

In a gaseous mixture at 4 atm pressure, 25% of molecules are Nitrogen, 40% of molecules are carbon dioxide and the rest are oxygen. The partial pressure of oxygen in the mixture is:

A
1.40 atm

B
1.6 atm

C
1 atm

D
0.9 atm

Solution

total pressure $$= 4atm $$
nitrogen $$= 25\% $$
$$Co_{2}=40\%$$
oxygen $$=35% $$
(portal pressure )$$O_{2}=$$ total pressure $$\times $$(mole tractra)$$O_{2}$$
=$$4\times \dfrac{35}{100}$$
=$$\dfrac{7}{5}$$
=$$1.40atm$$
Question 8
1 / -0

The vapour pressure is least for?

A
pure water

B
0.1m aqueous urea

C
0.2m aqueous urea

D
0.3m aqueous urea

Solution

Vapour pressure decreases as content of any impurity increases in solution.
$$\therefore V.P$$ will be least for $$0.3m$$ aq.urea
Question 9
1 / -0

How many grams of sucrose must be added to $$320g$$ of water to lower the vapour pressure by $$1.5mm$$ $$Hg$$ at $${25}^{o}C$$?
(Given: The vapour pressure of water at $${25}^{o}C$$ is $$23.8mm$$ $$Hg$$ and molar mass of sucrose is $$324.3g/mol$$)

A
$$21.5g$$

B
$$140g$$

C
$$363.36g$$

D
$$160.12g$$

Solution

$$w_{1}=$$ Wsucrose = ?
$$M_{1}=$$ Molecular weight $$+32,4.3\ g/mol$$
$$w_{2}=320\ g$$
$$M_{2}=H_{2}O=18\ g/mol$$
$$p-ps=1.5\ mm$$
$$p=23.8\ mm$$
$$\dfrac{p-ps}{p}=\dfrac{W_{1}}{M_{1}}\times \dfrac{M_{2}}{W_{2}}$$
$$\dfrac{1.5}{23.8}=\dfrac{W_{1}}{324.3}\times \dfrac{18}{320}$$
$$W_{1}=363.36\ g$$
Question 10
1 / -0

The vapour pressure of a liquid decreases by $$10 torr$$ when a non-volatile solute is dissolved. The mole fraction of the solute in solution is $$0.1$$. What would be the mole fraction of the liquid if the decrease in vapour pressure is $$20torr$$, the same solute being dissolved?

A
$$0.2$$

B
$$0.9$$

C
$$0.8$$

D
$$0.6$$

Solution

Answer is option $$C.$$
According to Rault's law,
$$\cfrac{P^0-P}{P}=X_a\longrightarrow (1)$$ Where $$X_a=\text{Mole fraction of solute.}$$
$$\underline {\text{Case 1}:}$$ $$P^0-P=10$$ $$torr,$$ $$X_a=0.1$$
$$\therefore$$ $$\cfrac{10}{P}=0.1$$ or $$P=\cfrac{10}{0.1}=100$$ $$torr$$
$$\underline {\text{Case 2}:}$$ $$P^0-P=20$$ $$torr,$$
$$\therefore (1)\longrightarrow \cfrac{20}{100}=X_a\Longrightarrow X_a=0.2$$
We have $$X_a+X_b=1,$$ where $$X_b=\text{Mole fraction of solvent.}$$
$$\therefore X_b=1-0.2=0.8$$

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States of Matter Test - 36

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States of Matter Test - 36

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Question 1

$$\dfrac {pV}{T}=\dfrac {2}{3}K$$

$$\dfrac {pV}{T}=\dfrac {3}{2}K$$

$$\dfrac {pV}{T}=2K$$

$$\dfrac {pV}{T}=3K$$

Solution

Question 2

$$K.E.$$ is dependent on temperature

$$K.E.$$ is dependent on pressure

The molecular collisions are perfectly elastic collisions

The pressure of the gas is due to collisions of gas molecules on the walls of the vessel

Solution

Question 3

$$1:4$$

$$7:32$$

$$1:8$$

$$3:16$$

Solution

Question 4

Phenol+ Aniline

Benzene + Toluene

Acetone + Chloroform

water + $$HCl$$

Solution

Question 5

$$NO_2$$

$$N_2O$$

$$N_2O_3$$

$$NO$$

$$CO_2$$

Solution

Question 6

$$1g$$

$$2g$$

$$6g$$

$$12g$$

Solution

Question 7

1.40 atm

1.6 atm

1 atm

0.9 atm

Solution

Question 8

pure water

0.1m aqueous urea

0.2m aqueous urea

0.3m aqueous urea

Solution

Question 9

$$21.5g$$

$$140g$$

$$363.36g$$

$$160.12g$$

Solution

Question 10

$$0.2$$

$$0.9$$

$$0.8$$

$$0.6$$

Solution

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