States of Matter Test

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States of Matter Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

A container contains equal mass of $$C{ O }_{ 2 },{ O }_{ 2 },C{ H }_{ 4 },S{ O }_{ 2 }$$ gases at a particular temperature.Which gas will exert the maximum amount of partial pressure?

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

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

C
$$C{ H }_{ 4 }$$

D
$$S{ O }_{ 2 }$$

Solution

Molecular weight of $$C{{O}_{2}}=12+16+16=44$$

Molecular weight of $${{O}_{2}}=16+16=32$$

Molecular weight of $$C{{H}_{4}}=12+4=16$$

Molecular weight of $$S{{O}_{2}}=32+16+16=64$$

So;

$$S{{O}_{2}}$$has maximum Molecular weight.

Hence $$S{{O}_{2}}$$ exert maximum amount of partial pressure.
Question 2
1 / -0

The kinetic energy for 14 grams of nitrogen gas at $$127^0C$$ is nearly (mol. mass of nitrogen = 28 and gas constant = 8.31 $$JK^{-1} mol^{-1}$$)

A
$$1230\ J$$

B
$$4015\ J$$

C
$$2493\ J$$

D
$$990\ J$$

Solution

$$KE=\cfrac {3}{2}nRT$$ $$n=\cfrac {14}{28}=0.5$$
$$=\cfrac {3}{2}\times 0.5 \times 8.314 \times (127+273)$$
$$=2494.2 J$$
Question 3
1 / -0

If number of molecules of $$H_{2}$$ are double than that of $$O_{2}$$, then ratio of mean kinetic energy per molecule of hydrogen to that of oxygen at 300 K is:

A
2:1

B
1:2

C
2:3

D
1:4

Solution

Mean Kinetic energy is given as, $$K.E=\cfrac{3}{2}nRT$$
$$n=$$ no. of molecules.
$$R=$$ Constant.
$$T=$$ Temperature.
For $$H_2$$,
$$K.E_{H_2}=\cfrac{3}{2}n_{H_2}RT\longrightarrow (1)$$
For $$O_2$$,
$$K.E_{O_2}=\cfrac{3}{2}n_{O_2}RT\longrightarrow (1)$$
On dividing equation $$(1)$$ by equation $$(2)$$ we get,
$$\cfrac{K.E_{H_2}}{K.E_{O_2}}=\cfrac{n_{H_2}}{n_{O_2}}$$
given $$n_{H_2}=2n_{O_2}$$,
$$\cfrac{K.E_{H_2}}{K.E_{O_2}}=\cfrac{n_{2O_2}}{n_{O_2}}=\cfrac{2}{1}$$
Thus the ration is $$2:1.$$
Question 4
1 / -0

An amount of 1.00 g of a gaseous compound of borbon and hydrogen occupies 0.820 liter at 1.00 atm and at $$3^{0}C$$. The compound is (R=0.0820 liter atm $$mole^{-10}K^{-1};$$ at. wt:H=1.0,B=10.8)

A
$$BH_{3}$$

B
$$B_{4}H_{10}$$

C
$$B_{2}H_{6}$$

D
$$B_{3}H_{12}$$

Solution

Given,
volume - $$0.820\ l$$
mass - $$1.00\ g$$
pressure - $$1.00\ atm$$
temperature - $${3^{\rm O}}C$$
Using ideal gas equation - $$PV = nRT$$
$$n$$ - number of mole = $$\dfrac{m}{M}$$
$$m$$ - given mass
$$M$$ - molar mass
$$PV = \dfrac{m}{M}RT$$
here, find the molar mass of the compound
$$M = \dfrac{{mRT}}{{PV}}$$
Firstly, change the temperature in Kelvin.
$${3^{\rm O}}C + 273 = 276K$$
Now, put the value in the formula.
$$M = \dfrac{{1 \times 0.0820 \times 276}}{{1 \times 0.820}}$$
= $$27.6\ g/mole$$
write each compound molar mass according to the given value and match it with this value.
(a) $$B{H_3}$$ - $$10.8 + 1 \times 3 = 13.8$$
(b) $${B_4}{H_{10}}$$ - $$4 \times 10.8 + 1 \times 10 = 53.2$$
(c) $${B_2}{H_6}$$ - $$2 \times 10.8 + 6 \times 1 = 27.6$$
(d) $${B_3}{H_{12}}$$ - $$3 \times 10.8 + 12 \times 1 = 44.4$$

So, option (c) is the correct answer.
Question 5
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Directions For Questions

When a vapour consisting of mixture of colourless gas X and a brown gas Y, at atmosphere pressure is gradually heated from $$25^oC$$, its colour is found to deepen at first and then to fade as the temperature is raised above $$160^oC$$. Atv$$600^oC$$, the vapour is almost colourless consisti=ng of a gas Z and $$O_2$$ but its colour deepens when the pressure is raised at this temperature.

...view full instructions

The gas Z in the above fraction is ?

A
Nitrous oxide

B
Nitric Oxide

C
Dinitrogen tetroxide

D
Nitrogen dioxide
Question 6
1 / -0

Equal mass which of the following substance of occupy maximum volume at room temperature?

A
Sugar

B
Water

C
Oxygen

D
Graphite

Solution

Lesser the molecular weight , the greater the substance occupies the volume
So here graphite has lower molecular weight. So It occupies more volume.
Option $$D$$ is correct.
Question 7
1 / -0

Normal boiling point of a liquid is that temperature at which vapour of the liquid pressure of the liquid is equal to:

A
zero

B
380mm of Hg

C
760mm of Hg

D
100mm of Hg

Solution
Question 8
1 / -0

Two liquids A and B have $${P}_{A}^{o}$$ and $${P}_{B}^{o}$$ in the ratio $$1:3$$ and the ratio of number of moles of $$A$$ and $$B$$ in liquid phase are $$1:3$$. Then mole fraction of A in vapour phase in equilibrium with the solution is equal to:

A
$$0.1$$

B
$$0.2$$

C
$$0.5$$

D
$$1.0$$

Solution

Partial pressure of $$A=K\times\cfrac{1}{4},$$ where $$K$$ is ratio constant.)
Partial pressure of $$B=3K\times \cfrac{3}{4}$$
Total pressure is $$\cfrac{10K}{4}$$.
Vapor phase mole fraction of $$A$$ is,
$$\cfrac{\text{Partial pressure of }A}{\text{Total pressure}}=\cfrac{K\times \cfrac{1}{4}}{\cfrac{10K}{4}}=\cfrac{1}{10}=0.1$$
Question 9
1 / -0

Correct gas equation is :

A
$$\dfrac{V_1T_2}{P_!}=\dfrac{V_2T_1}{P_2}$$

B
$$\dfrac{P_1V_1}{P_2V_2}=\dfrac{T_1}{T_2}$$

C
$$\dfrac{P_1T_2}{V_1}=\dfrac{P_2V_2}{T_2}$$

D
$$\dfrac{V_1V_2}{T_1T_2}=P_1P_2$$

Solution
Question 10
1 / -0

Two flask $$A$$ and $$B$$ of equal volumes maintained temperature 300 K and 700 K contain equal mass of $$He(g)$$ and $$N_{2}(g)$$ respectively. What is the ratio of total translation kinetic energy of gas in flask $$A$$ to that of flask $$B$$?

A
$$1:3$$

B
$$3:1$$

C
$$3:49$$

D
None of these

Solution

$$K.E = \dfrac{1}{2}m{v}^{2}$$

$$v = \sqrt{\dfrac{3RT}{M}}$$ , Its Given Masses of $${H}_{2}$$ and $${N}_{2}$$ are Equal .

Ratio of K.E would become $$\dfrac{{T}_{1}}{{M}_{1}} : \dfrac{{T}_{2}}{{M}_{2}}$$

Ratio of Total translation kinetic energy of Flask A to that of flask B = $$\dfrac{300}{4} : \dfrac{700}{28}$$ = $$3 :1$$

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States of Matter Test - 44

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States of Matter Test - 44

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SHARING IS CARING

Question 1

$$C{ O }_{ 2 }$$

$${ O }_{ 2 }$$

$$C{ H }_{ 4 }$$

$$S{ O }_{ 2 }$$

Solution

Question 2

$$1230\ J$$

$$4015\ J$$

$$2493\ J$$

$$990\ J$$

Solution

Question 3

2:1

1:2

2:3

1:4

Solution

Question 4

$$BH_{3}$$

$$B_{4}H_{10}$$

$$B_{2}H_{6}$$

$$B_{3}H_{12}$$

Solution

Question 5

Nitrous oxide

Nitric Oxide

Dinitrogen tetroxide

Nitrogen dioxide

Question 6

Sugar

Water

Oxygen

Graphite

Solution

Question 7

zero

380mm of Hg

760mm of Hg

100mm of Hg

Solution

Question 8

$$0.1$$

$$0.2$$

$$0.5$$

$$1.0$$

Solution

Question 9

$$\dfrac{V_1T_2}{P_!}=\dfrac{V_2T_1}{P_2}$$

$$\dfrac{P_1V_1}{P_2V_2}=\dfrac{T_1}{T_2}$$

$$\dfrac{P_1T_2}{V_1}=\dfrac{P_2V_2}{T_2}$$

$$\dfrac{V_1V_2}{T_1T_2}=P_1P_2$$

Solution

Question 10

$$1:3$$

$$3:1$$

$$3:49$$

None of these

Solution

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