States of Matter Test

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States of Matter Test - 22

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Question 1
1 / -0

Rain drops are spherical in shape because of

A
Surface tension

B
Capillary

C
Downward motion

D
Acceleration due to gravity

Solution

It require minimum surface area to make a spherical surface compared to any other shape so that it feels minimum surface tension and thus acquire minimum potential energy.
Question 2
1 / -0

The vapour pressure of pure benzene at $$88^oC$$ is 957 mm and that of toluene at the same temperature is 379.5 mm. Calculate the composition of benzene-toluene mixture boiling at $$88^oC$$.

A
$$x_{benzene}=0.66; \,x_{toluene}=0.34$$

B
$$x_{benzene}=0.34; \,x_{toluene}=0.66$$

C
$$x_{benzene} = x_{toluene}=0.5$$

D
$$x_{benzene}=0.75;\, x_{toluene}=0.25$$

Solution
Question 3
1 / -0

The specific heat of a bivalent metal is 0.16. The approximate equivalent mass of the metal will be:

A
40

B
20

C
80

D
10

Solution

atomic weight = $$\dfrac{6.4}{specific heat}$$

atomic weight = 40
equivalent weight = $$\dfrac{40}{2}$$ (bivalent)

equivalent mass = 20
Question 4
1 / -0

Four flasks of 1 litre capacity each arc separately filled with gases $$H_2, He, O_2$$ and $$O_3$$. At the same temperature and pressure the ratio of the number of atoms of these gases present in different flasks would be:

A
1 : 1 : 1 : 1

B
2 : 1 : 2 : 3

C
1 : 2 : 1 : 3

D
3 : 2 : 2 : 1

Solution

Ans ; A
Here in this question 4 different types of gasses are filled in same volume of flasks i.e. all 4 types of gasses have same number of molecules.
THe ratio of number of atoms of these gasses present in different flasks would be = 1:1:1:1
Question 5
1 / -0

In the reaction, $$N_2 + 3H_2 \rightarrow 2NH_3$$, the ratio of volumes of nitrogen, hydrogen and ammonia is 1 : 3: 2. These figures illustrate the law of:

A
constant proportions

B
Gay-Lussac

C
multiple proportions

D
reciprocal proportions

Solution

ANS; B
Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.
Question 6
1 / -0

At $$25^oC$$, the total pressure of an ideal solution obtained by mixing 3 mole of A and 2 mole of B, is 184 torr. What is the vapour pressure (in torr) of pure B at the same temperature (Vapour pressure of pure A at $$25^oC$$ is 200 torr.)?

A
180

B
160

C
16

D
100

Solution

$$x_A=\dfrac{n_A}{n_A+n_B}=\dfrac{3}{3+2}=\dfrac{3}{5}=0.6$$
$$x_B=0.4$$
$$p=p^0_A+p^0_Bx_B$$
$$184=200\times 0.6+p^0_B\times 0.4$$
$$184=120+p^0_B\times 0.4$$
$$p^0_B=\dfrac{64}{0.4}=160 torr$$
Question 7
1 / -0

Vapour pressure of pure $$A(p^0_A)=100$$ mm Hg
Vapour pressure of pure $$B(p^0_B)=150$$ mm Hg

2 mole of liquid A and 3 mole of liquid B are mixed to form an ideal solution. The vapour pressure of the solution will be :

A
135 mm

B
130 mm

C
140 mm

D
145 mm

Solution
Question 8
1 / -0

Equal volumes of different gases at any definite temperature and pressure have:

A
equal weights

B
equal masses

C
equal densities

D
equal number of moles

Solution

At equal volume of different gases at any definite temperature and pressure have equal no. of particles.
hence, equal number of moles.
Question 9
1 / -0

Mole fraction of the component A in vapour phase is $$x_1$$ and mole fraction of component A in liquid mixture is $$x_2$$, then ($$p^0_A =$$ vapour pressure of pure A; $$p^0_B =$$ vapour pressure of pure B), the total vapour pressure of liquid mixture is :

A
$$p^0_A\dfrac{x_2}{x_1}$$

B
$$p^0_A\dfrac{x_1}{x_2}$$

C
$$p^0_B\dfrac{x_1}{x_2}$$

D
$$p^0_B\dfrac{x_2}{x_1}$$

Solution

Mole fraction of component A in vapour phase is $$x_1$$
that means
$$P_A = P^0_A * x_1$$
and mole fraction of compound A in liquid mistureis $$x_2$$
that means we have,
$$P_A = x_2 * P^0_A$$
so, from both equation we get
$$x_2 * P^0_A = x_1 * P_t$$

so, $$P_t = \dfrac{x_2}{x_1}* P^0_A$$
Question 10
1 / -0

The following graphs illustrate:

A
Dalton's law

B
Boyle's law

C
Charles' law

D
Gay-Lussac's law

Solution

Chrle's law-
It states that at constant pressure, the volume of a fixed mass of a gas is directly proportional to the temperature.
$$PV=nRT\longrightarrow $$ Constant pressure
$$V=(\cfrac{nR}{P})T\Longrightarrow y=mx$$
$$V\propto T$$ straight line,

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States of Matter Test - 22

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States of Matter Test - 22

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Question 1

Surface tension

Capillary

Downward motion

Acceleration due to gravity

Solution

Question 2

$$x_{benzene}=0.66; \,x_{toluene}=0.34$$

$$x_{benzene}=0.34; \,x_{toluene}=0.66$$

$$x_{benzene} = x_{toluene}=0.5$$

$$x_{benzene}=0.75;\, x_{toluene}=0.25$$

Solution

Question 3

40

20

80

10

Solution

Question 4

1 : 1 : 1 : 1

2 : 1 : 2 : 3

1 : 2 : 1 : 3

3 : 2 : 2 : 1

Solution

Question 5

constant proportions

Gay-Lussac

multiple proportions

reciprocal proportions

Solution

Question 6

180

160

16

100

Solution

Question 7

135 mm

130 mm

140 mm

145 mm

Solution

Question 8

equal weights

equal masses

equal densities

equal number of moles

Solution

Question 9

$$p^0_A\dfrac{x_2}{x_1}$$

$$p^0_A\dfrac{x_1}{x_2}$$

$$p^0_B\dfrac{x_1}{x_2}$$

$$p^0_B\dfrac{x_2}{x_1}$$

Solution

Question 10

Dalton's law

Boyle's law

Charles' law

Gay-Lussac's law

Solution

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