Solutions Test 43

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Solutions Test 43

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Question 1
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Directions For Questions

Liquids A and B forms an ideal solution . At TK , $$ P^0_A = 0.4\, bar $$ and $$ P^0_B = 0.6\, bar $$. In a cylinder piston arrangement, 2.0 moles of vapours of liquid A and 3.0 moles of vapours of liquid B are taken at TK and 0.3 bar

...view full instructions

Which of the following is the only incorrect informarion regarding the composition of the system at 0.51 bar pressure ?

A
$$ X_A = 0.45 $$

B
$$ Y_A = \frac {6}{17} $$

C
$$ n_{A(liquid) } = \frac {12}{11} $$

D
$$ n_{A(vapour) } = \frac {12}{11} $$
Question 2
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Directions For Questions

Liquids A and B forms an ideal solution . At TK , $$ P^0_A = 0.4\, bar $$ and $$ P^0_B = 0.6\, bar $$. In a cylinder piston arrangement, 2.0 moles of vapours of liquid A and 3.0 moles of vapours of liquid B are taken at TK and 0.3 bar

...view full instructions

If the vapour are compressed slowly and isothermally , at what pressure , the first drop of liquid will appear ?

A
0.4 bar

B
0. 5 bar

C
0.52 bar

D
0.6 bar

Solution
Question 3
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One having high vapour pressure at temperature below its melting point is?

A
Benzoic acid

B
Salicylic acid

C
Citric acid

D
All of these

Solution

Benzoic acid appears as a white crystalline solid. Vapour pressures of benzoic acid were measured in the temperature range $$316\,K$$ to $$391 \,K$$ which is below its melting point $$395.1\, K$$. Hence correct answer is option A.
Question 4
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$$2.0\ g$$ mixture of sodium carbonate and sodium bicarbonate on heating to constant weight gave $$224\ mL$$ of $$CO_{2}$$ at N.T.P.
The $$\%$$ weight of sodium bicarbonate in the mixture is:

A
$$50$$

B
$$54$$

C
$$80$$

D
$$84$$

Solution

On heating the mixture only sodium bicarbonate dissociates. Sodium carbonate does not dissociate due to its stability.

$$2NaHCO_3 + heat \rightarrow Na_2CO_3 + CO_2 + H_2O$$

Let the weight of $$Na_2CO_3$$ in the mixture be $$x$$ g

Weight of $$NaHCO_3$$ in the mixture $$=(2-x)$$ g

No. of moles of $$Na_2CO_3$$ $$=\dfrac {x}{106}$$ moles

No. of moles of $$NaHCO_3$$ $$=\dfrac{(2-x)}{84}$$ moles

From the equation:
2 moles of $$NaHCO_3$$ give 1 mole of $$CO_2$$ gas

$$\dfrac{(2-x)}{84}$$ moles of $$NaHCO_3$$ give $$\dfrac{(2-x)}{84} \times \dfrac {1}{2} = \dfrac {(2-x)}{168}$$ moles of $$CO_2$$ gas

At N.T.P
22400 ml of $$CO_2$$ gas = 1 mole

224 ml of $$CO_2$$ gas = $$\dfrac{1}{22400} \times 224 = 0.01$$ moles

Now
$$\dfrac {(2-x)}{168} = \dfrac {1}{100}$$

$$\Rightarrow x = 0.32$$ g

Weight of sodium bicarbonate $$=(2-0.32) g = 1.68$$ g

Weight percentage of sodium bicarbonate = $$\dfrac{1.68}{2} \times 100 = 84$$ %
Question 5
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The compound $$Na_2 CO_3 . x H_2O$$ has $$50 \% H_2O$$ by mass. The value of "x" is :

A
4

B
5

C
6

D
7

E
8

Solution

(i) Molar mass of $$Na_2CO_3 = 106 $$ unit

Since mass % of the water in the compound is 50%.

$$\Rightarrow $$ Mass of the hydrated compound $$= 2\times 106=212g$$

i.e. Mass of water $$=106 u$$ per hydrated compound.

$$\therefore $$ Number of water molecules, $$x=\dfrac{mass}{molecular \ mass}=\dfrac{106}{18}\approx 6$$

Hence option C correct.
Question 6
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Directions For Questions

Liquid A ( molar mass = 100 g/mol and density = 1.25 g/ml ) and B ( molar mass = 50 g/mol and density = 1.0 g/ml) form an ideal solution . the molal volume of liquid A increases by afactor of 3800 as it vaporizes at $$ 27^0 $$C and that of liquid B increases by factor of 7600 at $$ 27^0 C $$ A solution of liquid A and B at $$ 27^0 C $$ has a vapour pressure of 54.0 (n R -= 0.08 atm/K-mol )

...view full instructions

The vapour pressure of pure liquid A at $$ 27^0 C $$ is

A
60 torr

B
48 torr

C
49.26 torr

D
61.58 torr

Solution
Question 7
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Directions For Questions

The lowering vapour pressure on adding non volatile solute in a pure liquid solvent may be measured by ostwald walker method. in this method dry air is first passed through a series of vessels having the solution, then through a series of vessels having pure solvent and finally through a vessel ( normally U Tube) having the absorbent of the solvent.the masses of the solution and pure solvent decreases due to removal of vapour of solvent in the flow of air and the m,ass of absorbent increases due to absorption of the vapour of solvent , in order to make the air, dery . by measuring the changes in mass we may determine the lowering of vapour pressure and hence, the composition of solution using raoults law.

...view full instructions

If the experimenty is performed aqueous $$ AlCl_3 $$ solution (a= 0.8 ) prepared by dissolving 1 mole of $$ AlCl_3 $$ in 17 mole of water and the decrease in the mass of solution in the experiment is found to be 0.18 g then the increase in the mass of absorbent should be

A
0.216 g

B
0.225 g

C
0.191 g

D
1.08 g

Solution

$$AlCl_3 \rightleftharpoons Al^{3+} +3Cl^{-}$$
$$1-0.8\\=0.2$$ $$0.8$$ $$3\times 0.8=\\2.4$$

Total no. of mole$$=0.2 + 0.8 +2.4 =3.4$$

Now, decrease in mass of solution $$\propto P$$
and increase in mass of absorber $$\propto P^o$$

$$\therefore \cfrac{P^o}{P}=X_2=\cfrac{17}{17+3.4}=\cfrac{5}{6}=\cfrac{0.18}{|\Delta m|_{absorber}}$$

$$\therefore$$ Increase in mass of absorber $$=0.216\ gm$$
Question 8
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At $$80^o C$$, the vapour pressure of pure liquid A is 520 mm Hg and that of pure liquid B is 1000 mm Hg. If a mixture solution of A and B boils at $$80^o C$$and 1 atm pressure, the amount of A in the mixture is :
(1 atm=760 mm Hg)

A
34 mol%

B
48 mol%

C
50 mol%

D
52mol%
Question 9
1 / -0

Sugar in urine sample can be detected by:

A
Fehling's solution

B
Benedict's solution

C
Tollen's reagent

D
all the three $$(a), (b)$$ and $$(c)$$

Solution
Question 10
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$$50 mL$$ of solution of $$ Na_2CO_3 $$ neutralizes $$49.35 mL$$ of $$4 N \ HCl$$ . the reaction is represented as
$$ CO^{2-}_3 +2H^+ \rightarrow CO_2 +H_2O $$
The density of this $$ Na_2CO_3 $$ solution is 1.25 g $$ mL^{-1} ,$$ the percentage of $$ Na_2CO_3$$ in it is :

A
$$47.7$$

B
$$37.7$$

C
$$26.7$$

D
$$16.7$$

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Solutions Test 43

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Solutions Test 43

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SHARING IS CARING

Question 1

$$ X_A = 0.45 $$

$$ Y_A = \frac {6}{17} $$

$$ n_{A(liquid) } = \frac {12}{11} $$

$$ n_{A(vapour) } = \frac {12}{11} $$

Question 2

0.4 bar

0. 5 bar

0.52 bar

0.6 bar

Solution

Question 3

Benzoic acid

Salicylic acid

Citric acid

All of these

Solution

Question 4

$$50$$

$$54$$

$$80$$

$$84$$

Solution

Question 5

4

5

6

7

8

Solution

Question 6

60 torr

48 torr

49.26 torr

61.58 torr

Solution

Question 7

0.216 g

0.225 g

0.191 g

1.08 g

Solution

Question 8

34 mol%

48 mol%

50 mol%

52mol%

Question 9

Fehling's solution

Benedict's solution

Tollen's reagent

all the three $$(a), (b)$$ and $$(c)$$

Solution

Question 10

$$47.7$$

$$37.7$$

$$26.7$$

$$16.7$$

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