States of Matter Test

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States of Matter Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

An LPG cylinder can withstand pressure difference of 15 atm across its boundaries. If at room temp $$({ 27 }^{ o })$$ it is filled with 2 atm pressure. Determine the temperature at which it will explode?

A
$${ 216 }^{ o }C$$

B
$${ 2127 }^{ o }C$$

C
$${ 2400 }^{ o }C$$

D
$${ 27 }^{ o }C$$

Solution

Given,
Pressure difference across LPG cylinder boundaries= 15 atm
We know atmospheric pressure = 1 atm
It is given that pressure difference across wall =15 atm
therefore,Pressure inside cylinder's wall=15+1=16 atm
$${P}_{1}=16atm$$
$${P}_{2}=2atm$$
$${T}_{2}=300K$$
Now according to gay lusaac's law at constant $$V$$
$$P\propto T$$
$$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } $$
$${T}_{1}=16 \times 300/2$$
$${T}_{1}=2400K$$
Question 2
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Vapour pressure of a saturated solution of a sparingly soluble salt $$A_2B_3$$ is 31.8 mm of Hg at $$40^{\circ}C$$. If vapour pressure of pure water is 31.8 mm of Hg at $$40^{\circ}C$$ the solubility product of $$A_2B_3$$ at $$40^{\circ}C$$ is:

A
$$5.67\times 10^{-6}$$

B
$$1.42\times 10^{-6}$$

C
$$6.30\times 10^{-5}$$

D
1

Solution
Question 3
1 / -0

Vapour pressure of pure benzene is 119 torr and of toluene is 37.0 torr at the same temperature mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be:

A
$$0.137$$

B
$$0.205$$

C
$$0.237$$

D
$$0.435$$
Question 4
1 / -0

$$18.0\ g$$ of glucose$$\left( {{C_6}{H_{12}}{O_6}} \right)$$ is added to $$178.2\ g$$ of water. The vapour pressure of water for this aqueous solution at $${100^o}\ C$$ is :

A
$$759..00$$ Torr

B
$$7.60$$ Torr

C
$$76.00$$ Torr

D
$$752.40$$ Torr

Solution

Molecular mass of water $$=2\times 1+1\times 16=18$$g
For $$178.2g$$ water $$n_A$$=9.9
Molecular mass of glucose $$=6\times 12+12\times 1+6\times 16=180$$g
For 18g glucose $$n_B$$=0.1
$$X_B=\dfrac{0.10}{(0.1+9.9)}=0.01$$
$$X_A=0.99$$
For lowering of vapour pressure ,
$$p=p^0_AX^A=P^0_A(1−X_B)$$
$$P=760(1−0.01)$$
$$=760−7.6$$
$$=752.4torr$$
Question 5
1 / -0

100 g of liquid $$A$$ (molar mass 140 g $$mol^{-1}$$) was dissolved in 1000 g of liquid $$B$$ (molar mass 180 g $$mol^{-1}$$) The vapour pressure of pure liquid $$B$$ was found to be 500 torr.

Calculate the vapour pressure of pure liquid $$A$$ and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr:

A
$$3.19, 500$$

B
$$283.18, 31.999$$

C
$$500, 300$$

D
$$190, 31.965$$

Solution

Given: $$P_T= 475, W_A= 100g, M_A=140, {P^o_B}=500$$
$$W_B=1000g, M_B= 180, P^o_A ?, P_A=?$$
$$P_T=P^o_AX_A+P^o_BX_B\longrightarrow (1)$$ & $$X_A+X_B=1 \longrightarrow (2)$$
$$\Rightarrow \cfrac {X_B}{X_A}=\cfrac {W_BM_A}{W_AM_B}=\cfrac {1000\times 140}{100\times 180}=7.78$$
$$\therefore X_B= 7.78 X_A$$
Substituting $$X_B$$ in $$(2)$$
$$X_A+7.78X_A=1$$
$$\therefore X_A=0.113$$ & hence $$X_B=1-0.113=0.886$$
Substituting $$X_A$$ & $$X_B$$ in equation $$(1)$$
$$\Rightarrow 475= P^o_A(0.113)+500(0.886)$$
$$\Rightarrow 0.113P^o_A=32 \Rightarrow P^o_A=\cfrac {32}{0.113}=283.18 torr$$
Now, $$P_A=P^o_A(1-X_B)=P^o_AX_A$$
$$= 283.18 \times 0.113$$
$$= 31.999 torr$$
Question 6
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The vapour pressure of a pure liquid $$A$$ is $$70$$ torr at $$27^oC$$. It forms an ideal solution with another liquid $$B$$. The mole fraction of $$B$$ is $$0.2$$ and total vapour pressure of the solution is $$84$$ torr at $$27^oC$$. The vapour pressure of pure liquid $$B$$ at $$27^o$$C is:

A
$$14$$

B
$$56$$

C
$$140$$

D
$$70$$

Solution

We know Raoult's law,

For an ideal solution,

Partial pressure in solution $$=$$ mole fraction $$\times$$ Vapour pressure in pure state

Given,

$${ P }_{ A }^{ 0 }=70$$ torr at $${ 27 }^{ 0 }C$$ $${ X }_{ B }=0.2$$

then $${ X }_{ A }=1-{ X }_{ B }=1-0.2=0.8$$ $${ P }_{ B }^{ 0 }=?$$

$${ P }_{ total }$$ or $${ P }_{ solution }$$ $$=84$$ torr

By, Raoult's law

$$\boxed { { P }_{ solution }={ X }_{ A }{ P }_{ A }^{ 0 }+{ X }_{ B }{ P }_{ B }^{ 0 } } $$

$$84=0.8\times 70+0.2{ P }_{ B }^{ 0 }$$

$${ P }_{ B }^{ 0 }=\dfrac { 84-56 }{ 0.2 } $$

$${ P }_{ B }^{ 0 }=140$$ torr

$$\therefore$$ Vapour pressure of pure liquid $$B$$ at $${ 25 }^{ 0 }C$$ is $$140$$ torr.

Hence, the correct option is $$\text{C}$$
Question 7
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Which of the following change is observed occurs when a substance X is converted from liquid to vapour phase at the standard boiling point?
I.Potential energy of the system decreases
II. The distance between molecules increases
III. The average kinetic energy of the molecules in both phases are equal.

A
I only

B
II only

C
III only

D
II and III only

Solution
Question 8
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$${ \Delta }_{ f }{ G }^{ }$$ at 500 K for substance $$S$$ in liquid state and gaseous state are +100.7 kcal $${ }^{ -1 }$$ and +103 kcal $${ }^{ -1 }$$, respectively. Vapour pressure of liquid $$S$$ at 500 K is approximately equal to:
($$R=2\ cal $$ $${ K }^{ -1 }{ }^{ -1 }$$)

A
0.1 atm

B
10 atm

C
100 atm

D
1 atm

Solution

Solution:- (B) $$10 \; atm$$
As we know that, at equilibrium,
$${\Delta{G}}^{°} = -2.303 \; RT \; \log{{K}_{p}} ..... \left( 1 \right)$$
$${S}_{\left( l \right)} \rightleftharpoons {S}_{\left( g \right)}$$
From the above reaction,
$${\Delta{G}}^{°} = {{\Delta{G}}_{f}}_{\left( \text{product} \right)} - {{\Delta{G}}^{°}}_{\left( \text{reactant} \right)}$$
$$\Rightarrow {\Delta{G}}^{°} = 103 - 100.7 = 2.3 \; {kcal}/{mol} = 2.3 \times {10}^{3} \; {kcal}/{mol}$$
Given:-
$$T = 500 \; K$$
$$R = 2 {cal}/{mol-K}$$
From $${eq}^{n} \left( 1 \right)$$, we have
$$2.3 \times {10}^{3} = -2.303 \times 2 \times 500 \times \log{{K}_{p}}$$
$$\Rightarrow \log{{K}_{p}} = -1$$
$$\Rightarrow {K}_{p} = {10}^{-1} \; atm$$
Now, from the above reaction,
$${K}_{p} = \cfrac{1}{{P}_{{S}_{\left( l \right)}}}$$
$$\Rightarrow {P}_{{S}_{\left( l \right)}} = \cfrac{1}{{10}^{-1}} = 10 \; atm$$
Hence the vapou pressure of liquid $$S$$ is approximately equal to $$10 \; atm$$.
Question 9
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The vapour pressure of pure $$A$$ is 10 torr and at the same temperature when 1 g of $$B$$ is dissolved in 20 gm of $$A$$, its vapour pressure is reduced to 9.0 torn If the molecular mass of $$A$$ is 200 amu, then the molecular mass of $$B$$ is:

A
100 amu

B
90 amu

C
75 amu

D
120 amu

Solution

Given data

Wt of $$A=20$$ gm
$${ P }_{ A }^{ 0 }=10$$ torr
molecular mass of $$A=200$$ amu
Wt of $$B=1$$ gm
$${ P }_{ solution }=9$$ torr
molecular mass of $$B=?$$

This problem can be solved by using the relative lowering of vapour pressure.

$$\dfrac { { P }^{ 0 }-P }{ { P }^{ 0 } } =\dfrac { { n }_{ 2 } }{ { n }_{ 1 }+{ n }_{ 2 } } $$

$${ P }^{ 0 }=$$ Vapour pressure of pure solvent
$$P=$$ Vapour pressure of solution
$${ n }_{ 2 }=$$ moles of solute
$${ n }_{ 1 }=$$ moles of solvent

$$\dfrac { 10-9 }{ 10 } =\dfrac { \dfrac { 1 }{ x } }{ \dfrac { 20 }{ 200 } +\dfrac { 1 }{ x } } $$

$$\left[ \dfrac { 1 }{ 10 } +\dfrac { 1 }{ x } \right] \dfrac { 1 }{ 10 } =\dfrac { 1 }{ x } $$

$$\dfrac { 1 }{ 100 } +\dfrac { 1 }{ 10x } =\dfrac { 1 }{ x } $$

$$\dfrac { 1 }{ 100 } =\dfrac { 1 }{ x } -\dfrac { 1 }{ 10x } =\dfrac { 9 }{ 10x } $$

   $$\therefore$$ The molecular mass of compound (solute) B is $$90$$ amu

$$10x=900$$

$$\boxed { x=90amu } $$

Hence, the correct option is $$B$$
Question 10
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Pressure of $$COCl_2(g)$$ in the gaseous mixture in final state ?

A
$$80$$

B
$$\dfrac{80}{3}$$

C
$$\dfrac{160}{3}$$

D
$$\dfrac{40}{3}$$

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 63

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States of Matter Test - 63

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SHARING IS CARING

Question 1

$${ 216 }^{ o }C$$

$${ 2127 }^{ o }C$$

$${ 2400 }^{ o }C$$

$${ 27 }^{ o }C$$

Solution

Question 2

$$5.67\times 10^{-6}$$

$$1.42\times 10^{-6}$$

$$6.30\times 10^{-5}$$

1

Solution

Question 3

$$0.137$$

$$0.205$$

$$0.237$$

$$0.435$$

Question 4

$$759..00$$ Torr

$$7.60$$ Torr

$$76.00$$ Torr

$$752.40$$ Torr

Solution

Question 5

$$3.19, 500$$

$$283.18, 31.999$$

$$500, 300$$

$$190, 31.965$$

Solution

Question 6

$$14$$

$$56$$

$$140$$

$$70$$

Solution

Question 7

I only

II only

III only

II and III only

Solution

Question 8

0.1 atm

10 atm

100 atm

1 atm

Solution

Question 9

100 amu

90 amu

75 amu

120 amu

Solution

Question 10

$$80$$

$$\dfrac{80}{3}$$

$$\dfrac{160}{3}$$

$$\dfrac{40}{3}$$

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