Solutions Test 44

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Solutions Test 44

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Weekly Quiz Competition

Question 1
1 / -0

Dry air is bubbled successively through (i) a solution, (ii) its solvent, and (iii) through $$CaCl_2$$. The lowering of vapour pressure of the solvent due to the addition of solute is equal to

A
$$p^0 \times \dfrac {loss\ in\ wt. of\ solvent}{gain\ in\ wt\ of\ CaCl_2}$$

B
$$\dfrac {loss\ in\ wt. of\ solvent}{gain\ in\ wt\ of\ CaCl_2}$$

C
$$\dfrac {loss\ in\ wt. of\ solution}{gain\ in\ wt\ of\ CaCl_2}$$

D
$$\dfrac {loss\ in\ wt. of\ solute}{gain\ in\ wt\ of\ CaCl_2}$$

Solution
Question 2
1 / -0

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 3
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 4
1 / -0

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 5
1 / -0

On mixing heptane and octane from an ideal solution. At $$373 K$$, the vapour pressure of the two liquid compounds (heptane and octane) are $$105\ kPa$$ and $$45\ kPa$$ respectively. Vapour pressure of the solution obtained by mixing $$250g$$ of heptane and $$35g$$ of octane will be (molar mass of heptane $$=100$$ and octane $$=114\ g\ mol^{-1}$$)

A
$$72.0\ kPa$$

B
$$36.1\ kPa$$

C
$$98.4\ kPa$$

D
$$114.5\ kPa$$

Solution

Moles of Heptane = $$\mathrm{\cfrac{250}{100} =2.5}$$

Moles of Octane = $$\mathrm{\cfrac{35}{114} = 0.3}$$

Mole fraction of Heptane = $$\mathrm{\cfrac{2.5}{2.8} = 0.89}$$ and that of Octane = $$\mathrm{\cfrac{0.3}{2.8} = 0.11}$$

According to Raoult's Law:
$$\mathrm{P_T =P^0_A X_A +P^0_BX_B}$$

$$\mathrm{\implies P_T = (0.89 \times 105 + 0.11 \times 45)}$$ kPa

$$\mathrm{\implies P_T = 98.4}$$ kPa
Question 6
1 / -0

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 7
1 / -0

Which of the following is true when components forming an ideal solution are mixed?

A
$$\Delta H_m = \Delta V_m = 0$$

B
$$\Delta H_m > \Delta V_m$$

C
$$\Delta H_m < \Delta V_m$$

D
$$\Delta H_m = \Delta V_m = 1$$

Solution

For the ideal solution $$\Delta H_{mix}$$ and $$\Delta V_{mix} = 0.$$
Question 8
1 / -0

Which pair from the following will not form an ideal solution?

A
$$CCl_4 + SiCl_4$$

B
$$H_2O + C_4H_9OH$$

C
$$C_2H_5Br + C_2H_5I$$

D
$$C_6H_{14} + C_7H_{16}$$

Solution

$$H_2O$$ and $$C_4H_9OH$$ do not form ideal solution because there is hydrogen bonding between $$H_2O$$ and $$C_4H_9OH.$$
Question 9
1 / -0

Which of the following mixture shows positive deviation by ideal behaviour

A
$$CHCl_3 + (CH_3)_2CO$$

B
$$C_6H_6 + C_6H_5CH_3$$

C
$$H_2O + HCl$$

D
$$CCl_4 + CHCl_3$$

Solution

$$A. CHCl_3+ (CH_3)_2O$$
Presence of strong H-bonding leading negative deviation from Raoult's law.
$$(B). C_6H_6+ C_6H_5CH_3\Rightarrow ideal \ solution.$$
$$(C). H_2O+HCl \Rightarrow$$ Water and hydrochloric acid form miscible solutions. They show no deviation
$$(D) CCl_4+CHCl_3$$ due to presence of Difference in polarity they exhibit +ve deviation from Raoult's law
Question 10
1 / -0

$$60 \,gm$$ of Urea (Mol. wt 60) was dissolved in $$9.9$$ moles of water. If the vapour pressure of pure water is $$P_0$$ , the vapour pressure of solution is

A
$$0.10 \,P_0$$

B
$$1.10 \,P_0$$

C
$$0.90 \,P_0$$

D
$$0.99 \,P_0$$

Solution

Given that,

Moles of water. $$N=9.9$$ moles

Moles of urea, $$n=\dfrac{w}{M}=\dfrac{60}{60}=1$$

Now,

$$\dfrac {P^{0} - P_{S}}{P^{0}} = \dfrac {n}{N} \Rightarrow \dfrac {P^{0} - P_{S}}{P^{0}} = \dfrac {1}{9.9} \Rightarrow 9.9 \,P^{0} - 9.9 \,P_s = P^{0}$$

$$\Rightarrow 8.9 \,P^{0} = 9.9 \,P_s \Rightarrow P_s = \dfrac {8.9}{9.9} P^{0} \approx 0.90 \,P^{0}$$

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Solutions Test 44

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Solutions Test 44

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Question 1

$$p^0 \times \dfrac {loss\ in\ wt. of\ solvent}{gain\ in\ wt\ of\ CaCl_2}$$

$$\dfrac {loss\ in\ wt. of\ solvent}{gain\ in\ wt\ of\ CaCl_2}$$

$$\dfrac {loss\ in\ wt. of\ solution}{gain\ in\ wt\ of\ CaCl_2}$$

$$\dfrac {loss\ in\ wt. of\ solute}{gain\ in\ wt\ of\ CaCl_2}$$

Solution

Question 2

500

600

700

None

Solution

Question 3

501

457.14

425

525

Solution

Question 4

499

399

299

none

Solution

Question 5

$$72.0\ kPa$$

$$36.1\ kPa$$

$$98.4\ kPa$$

$$114.5\ kPa$$

Solution

Question 6

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 7

$$\Delta H_m = \Delta V_m = 0$$

$$\Delta H_m > \Delta V_m$$

$$\Delta H_m < \Delta V_m$$

$$\Delta H_m = \Delta V_m = 1$$

Solution

Question 8

$$CCl_4 + SiCl_4$$

$$H_2O + C_4H_9OH$$

$$C_2H_5Br + C_2H_5I$$

$$C_6H_{14} + C_7H_{16}$$

Solution

Question 9

$$CHCl_3 + (CH_3)_2CO$$

$$C_6H_6 + C_6H_5CH_3$$

$$H_2O + HCl$$

$$CCl_4 + CHCl_3$$

Solution

Question 10

$$0.10 \,P_0$$

$$1.10 \,P_0$$

$$0.90 \,P_0$$

$$0.99 \,P_0$$

Solution

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