States of Matter Test

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States of Matter Test - 75

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Question 1
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Directions For Questions

The pressure of two pure liquid A and B which form an ideal solutions are 400 mm Hg and 800 mm Hg respectively at temperature T. A liquid containing 3 : 1 molar composition pressure can be varied. The solutions is slowly vaporized at temperature T by decreasing the applied pressure starting with a pressure of 760 mmHg. A pressure gauge (in mm) Hg is connected which given the reading of pressure applied.

...view full instructions

The reading of pressure gauge at bubble point is :

A
500

B
600

C
700

D
None

Solution

$$ X_A = 0.75 \quad X_B = 0.05 $$ above 500 mm Hg only liquid phase exists .
$$ P_{bubble} point = X_A P^0_A + X_B P^0_B $$
$$ P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B $$
$$ 0.75 +400 + 0.25 \times 800 = 500 mm $$
$$ y_A = 0.75 \quad \quad y_B = 0.5 $$
at dew point
$$ \frac {1}{P_T } = \frac {Y_A }{P^0A} + \frac {Y_B}{P^0_B } \Rightarrow \frac {1}{ P_T} = \frac { 0.75}{400} + \frac { 0.25}{800} = \frac { 1.5 + 0.25}{800} $$
$$ \Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg $$
Below dew point only vapor phase exists.
Question 2
1 / -0

The reading of pressure gauge at which only vapour phase exists is

A
501

B
457.14

C
425

D
525

Solution

$$ X_A = 0.75 \quad X_B = 0.05 $$ above 500 mm Hg only liquid phase exists .
$$ P_{bubble} point = X_A P^0_A + X_B P^0_B $$
$$ P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B $$
$$ 0.75 +400 + 0.25 \times 800 = 500 mm $$
$$ y_A = 0.75 \quad \quad y_B = 0.5 $$
at dew point
$$ \frac {1}{P_T } = \frac {Y_A }{P^0A} + \frac {Y_B}{P^0_B } \Rightarrow \frac {1}{ P_T} = \frac { 0.75}{400} + \frac { 0.25}{800} = \frac { 1.5 + 0.25}{800} $$
$$ \Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg $$
Below dew point only vapor phase exists.
Question 3
1 / -0

The vapour pressure lowering caused by the addition of $$100 \,g$$ of sucrose (molecular mass = 342) to $$1000 \,g$$ of water if the vapour pressure of pure water at $$25 ^\circ C$$ is $$23.8 \,mm \,Hg$$

A
$$1.25 \,mm \,Hg$$

B
$$0.125 \,mm \,Hg$$

C
$$1.15 \,mm \,Hg$$

D
$$00.12 \,mm \,Hg$$

Solution

Given molecular mass of sucrose $$= 342$$

Moles of sucrose $$= \dfrac {100}{342} = 0.292 \,mole$$

Moles of water $$N = \dfrac {1000}{18} = 55.5 \,moles $$ and

Vapour pressure of pure water $$P^{0} = 23.8 \,mm \,Hg$$

According to Raoult's law

$$\dfrac {\Delta P}{P^{0}} = \dfrac {n}{n + N} \Rightarrow \dfrac {\Delta P}{23.8} = \dfrac {0.292}{0.292 + 55.5}$$

$$\Delta P = \dfrac {23.8 \times 0.292}{55.792} = 0.125 \,mm \,Hg$$
Question 4
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$$5 \,cm^3$$ of acetone is added to $$100 \,cm^3$$ of water, the vapour pressure of water over the solution

A
It will be equal to the vapour pressure of pure water

B
It will be less than the vapour pressure of pure water

C
It will be greater than the vapour pressure of pure water

D
It will be very large

Solution

Acetone solution has a vapour pressure less than pure water since the number of water molecules exposed to the surface will be less after mixing.
Question 5
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Directions For Questions

The pressure of two pure liquid A and B which form an ideal solutions are 400 mm Hg and 800 mm Hg respectively at temperature T. A liquid containing 3 : 1 molar composition pressure can be varied. The solutions is slowly vaporized at temperature T by decreasing the applied pressure starting with a pressure of 760 mmHg. A pressure gauge (in mm) Hg is connected which given the reading of pressure applied.

...view full instructions

The reading of pressure gauge at which only liquid phase exists

A
499

B
399

C
299

D
none

Solution

$$ X_A = \dfrac34=0.75; \quad X_B = 0.25 $$

$$ P_{\text{bubble point}} = X_A P^0_A + X_B P^0_B $$

$$ P_{\text{bubble point}} = X_A P^0_A + X_B P^0_B $$

$$ 0.75 +400 + 0.25 \times 800 = 500 mm $$

Above 500 mm Hg only liquid phase exists.

$$ y_A = 0.75; \quad \quad y_B = 0.5 $$

At dew point:

$$ \dfrac {1}{P_T } = \dfrac {Y_A }{P^0A} + \dfrac {Y_B}{P^0_B } \Rightarrow \dfrac {1}{ P_T} = \dfrac { 0.75}{400} + \dfrac { 0.25}{800} = \dfrac { 1.5 + 0.25}{800} $$

$$ \Rightarrow P_T = \dfrac {800}{1.75} = 457.14\ mm Hg $$

Below the dew point, only the vapour phase exists.
Question 6
1 / -0

If the pressure of the gas contained in a closed vessel is increased by $$ 20 $$ % when heated by $$ 273^{\circ} \,C $$ then it's initial temperature must have been:

A
$$ 1092^{\circ} \,C $$

B
$$ 1092\,K $$

C
$$ 1365^{\circ}\,C $$

D
$$ 1365\,K $$

Solution

$$ \dfrac{P}{t + 273} = \dfrac{1.2P}{t + 273 + 273} $$

$$ 1.2t + 273 \times 1.2 = t + 273 \times 2 $$ $$ \Rightarrow $$ $$ 0.2t = 273 \times 0.8 $$ $$ t = 273 \times 4 =1092K $$

Option B is correct.
Question 7
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Vapour pressure of a solution is

A
Directly proportional to the mole fraction of the solvent

B
Inversely proportional to the mole fraction of the solute

C
Inversely proportional to the mole fraction of the solvent

D
Directly proportional to the mole fraction of the solute

Solution

For solutions containing non - volatile solutes , the Raoult's law may be stated as at a given temperature, the vapour pressure of a solution containing non - volatile solute is directly proportional to the mole fraction of the solvent.
Question 8
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Which of the following pairs will diffuse at the same rate through a porous plug?

A
$$CO, NO_{2}$$

B
$$NO_{2}, CO_{2}$$

C
$$NH_{3}, PH_{3}$$

D
$$NOC_{2} H_{6}$$

Solution

Because both $$NO$$ and $$C_{2}H_{6}$$ have same molecular weights.

$$\left [ M_{NO}=M_{C_{2}H_{6}}=30 \right ]$$ and rate of diffusion $$\propto$$ molecular weight.
Question 9
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Which of the following solution in water possesses the lowest vapour pressure?

A
$$0.1 (M) NaCl$$

B
$$0.1 (M) BaCl_2$$

C
$$0.1 (M) KCl$$

D
None of these

Solution

The molarity for all the options is same i.e. 0.1 M, thus the deciding factor would be 'i'
in A, i = 2
in B, i = 3
in C, i = 2

This implies that dissociation into ions is maximum in the case of B, and thus least vapour pressure is observed
Question 10
1 / -0

To which of the following gaseous mixtures is Dalton's law not applicable?

A
$$Ne + He + SO_{2}$$

B
$$NH_{3} + HCl + HBr$$

C
$$O_{2} + N_{2} + CO_{2}$$

D
$$N_{2} + H_{2} + O_{2}$$

Solution

$$NH_{3}$$ and $$HCl$$ & $$HBr$$ is a reacting gas mixture to produce $$NH_{4}Cl$$ & $$NH_{4}Br$$.

Therefore, Dalton's law is not applicable.

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 75

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States of Matter Test - 75

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Rank

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SHARING IS CARING

Question 1

500

600

700

None

Solution

Question 2

501

457.14

425

525

Solution

Question 3

$$1.25 \,mm \,Hg$$

$$0.125 \,mm \,Hg$$

$$1.15 \,mm \,Hg$$

$$00.12 \,mm \,Hg$$

Solution

Question 4

It will be equal to the vapour pressure of pure water

It will be less than the vapour pressure of pure water

It will be greater than the vapour pressure of pure water

It will be very large

Solution

Question 5

499

399

299

none

Solution

Question 6

$$ 1092^{\circ} \,C $$

$$ 1092\,K $$

$$ 1365^{\circ}\,C $$

$$ 1365\,K $$

Solution

Question 7

Directly proportional to the mole fraction of the solvent

Inversely proportional to the mole fraction of the solute

Inversely proportional to the mole fraction of the solvent

Directly proportional to the mole fraction of the solute

Solution

Question 8

$$CO, NO_{2}$$

$$NO_{2}, CO_{2}$$

$$NH_{3}, PH_{3}$$

$$NOC_{2} H_{6}$$

Solution

Question 9

$$0.1 (M) NaCl$$

$$0.1 (M) BaCl_2$$

$$0.1 (M) KCl$$

None of these

Solution

Question 10

$$Ne + He + SO_{2}$$

$$NH_{3} + HCl + HBr$$

$$O_{2} + N_{2} + CO_{2}$$

$$N_{2} + H_{2} + O_{2}$$

Solution

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