States of Matter Test

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States of Matter Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

A $$4.40\ g$$ piece of solid $$CO_{2}$$ is allowed to subline in a ballon. The final volume of the ballon is $$1.00\ L$$ at $$300\ K$$. What is the pressure of gas?

A
$$0.122$$

B
$$2.46$$

C
$$1.22$$

D
$$24.6$$

Solution

$$moles\quad of\quad { CO }_{ 2 }=\frac { 4.40 }{ 44 } =0.1$$

$$ P=\frac { 0.1\times 0.0821\times 300 }{ 1 } $$

$$ =2.46V$$

conclusion: hence the answer option (B) is correct.
Question 2
1 / -0

The vapour pressure of water at $$20^{o}C$$ is $$17.54mm Hg$$. Then the vapour pressure of the water in the apparatus is lowered, decreasing the volume of the gas above the liquid to one half of its initial volume (temp. constant) is:

A
$$5.77\ mmHg$$

B
$$16\ mmHg$$

C
$$35.08\ mmHg$$

D
between $$8.77$$ and $$15.54\ mmHg$$

Solution

$We know that,$
$Mole fraction of sucrose: $$\dfrac {\text{ Moles of sucrose} }{\text{ Moles of sucrose}+\text{ Moles of water} } $$$
So,
$$X_{ sucrose }=\dfrac { \dfrac { 200g }{ 342.13gmol^{ -1 } } }{ \dfrac { 200g }{ 342.13gmol^{ -1 } } +\dfrac { 112.3g }{ 18.01gmol^{ -1 } } } =0.0857$$

Mole fraction of water:

$$X_{ water }=\dfrac { \dfrac { 112.3g }{ 18.01gmol^{ -1 } } }{ \dfrac { 112.3g }{ 18.01gmol^{ -1 } } +\dfrac { 200g }{ 342.13gmol^{ -1 } } } =0.914$$

Note that by definition, $$x_{ sucrose }+x_{ water }=1$$

Because sucrose is involatile, the vapour pressure of the solution is proportional to the mole fraction of water:
$$P_{ solution }=X_{ water }+17.54\ mm$$ of $$Hg$$
$$=0.914\times 17.54\ mm$$ of $$Hg=16\ mm$$ of $$Hg$$
Conclusion: hence the correct answer is not mentioned in the above options.
Question 3
1 / -0

For the equilibrium ,$${ 2SO }_{ 3 }(g)\rightleftharpoons 2{ SO }_{ 2 }(g)+{ O }_{ 2 }(g)$$; the partial pressure of $${ SO }_{ 3 }$$, $${ SO }_{ 2 }$$ and $${ O }_{ 2 }$$ gases at $$650 K$$ are respectively $$0.2 bar,0.6 bar$$ and $$0.4 bar$$. If the moles of both oxides of sulphur are so adjusted as equal, what will be the partial pressure of $${ O }_{ 2 }$$?

A
$$4.4$$

B
$$0.44$$

C
$$3.6$$

D
$$1.12$$

Solution

$$2 SO_3 \rightleftharpoons 2SO_2 + O_2$$
$$K_p = \dfrac{[SO_2]^2 [O_2]}{[SO_3]^2} = \dfrac{(0.6)^2 (0.4)}{(0.2)^2} = 3.6$$
If $$SO_2$$ and $$SO_3$$ are made equal,
$$3.6 = \dfrac{[SO_2]^2}{[SO_3]^2} \times [O_2] \Rightarrow [O_2] = 3.6$$
Because moles of $$SO_2 \, \& \, SO_3$$ are equal, $$P_{SO_2}, \, \& \, P_{SO_3}$$ came each other
Question 4
1 / -0

If $$P, V, M, T$$ and $$R$$ are pressure, volume, molar mass, temperature and gas constant respectively, then for an ideal gas, the density is given by:

A
$$\dfrac{RT}{PM}$$

B
$$\dfrac{P}{RT}$$

C
$$\dfrac{M}{V}$$

D
$$\dfrac{PM}{RT}$$

Solution

For ideal gas
PV=RT
V=RT/P
Density is given by
D=M/V
Where M is mass and V is volume
Putting the value of V
D=$$\frac { M }{ RT/P } $$
D=$$\frac { MP }{ RT } $$
Question 5
1 / -0

The volume of $$2.8\ g$$ of carbon monoxide at $$27^{o}C$$ and $$0.821 atm$$ pressure is:

A
$$0.03\ L$$

B
$$3\ L$$

C
$$0.3\ L$$

D
$$1.5\ L$$

Solution

The volume of carbon monoxide=$$\dfrac { nRT }{ P }$$
Mass of carbon\quad monoxide$$=2.8g$$
$$P=0.821\ atm,$$
$$T=27+273=300K$$
$$R=0.0821$$
because gram molecular weight of $$CO=12+16=28$$
because number of moles in 2.8g of $$CO=\dfrac { 2.8 }{ 28 } =0.1$$
therefore volume$$=\dfrac { 0.1\times 0.0821\times 300 }{ 0.821 } =3\ litre$$
Question 6
1 / -0

$$3$$ moles of $$P$$ and $$2$$ moles of $$Q$$ are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are $$80$$ and $$60$$ respectively?

A
$$80$$ torr

B
$$140$$ torr

C
$$720$$ torr

D
$$70$$ torr

Solution

Mole fraction of $$P=\dfrac { 3 }{ 3+2 } =\dfrac { 3 }{ 5 } $$

Mole fraction of $$Q=\dfrac { 2 }{ 3+2 } =\dfrac { 2 }{ 5 } $$

Hence,

Total vapour pressure =(Mole fraction of $$P$$ $$\times $$ Vapour pressure of $$P$$) + (mole fraction of $$Q$$ $$\times$$ Vapour pressure of $$Q$$)

$$=\left( \dfrac { 3 }{ 5 } ×80 \right) +\left( \dfrac { 2 }{ 5 } ×60 \right) =48+24=72\ torr$$
Question 7
1 / -0

Combination that does not obey Dalton's law:
$$A=CO$$
$$B=Cl_{2}$$
$$C=F_{2}$$
$$D=Xe$$

A
$$A,B$$

B
$$B,C$$

C
$$B,D$$

D
$$A,C$$

Solution

Dalton's law of partial pressure is not applicable to a combination of $$CO$$ and $$Cl_2$$ because they are reacting gases means they react with each other combines:
$${ CO }+{ Cl }_{ 2 }\longrightarrow { COCl }_{ 2 }$$
poisonous gas
Question 8
1 / -0

$$4$$ moles $$CO, 5$$ mole $$N_{2}$$ and $$2$$ mole $$C_{2}H_{4}$$ are placed in a $$5$$ litre at $$27^{o}C$$. The ratio of kinetic energy per molecular of $$CO_{2}$$, $$He$$ and $$NH_{3}$$ is:

A
$$4:5:2$$

B
$$5:5:4$$

C
$$2:4:5$$

D
$$1:1:1$$

Solution

Kinetic energy$$=\dfrac{3}{2}nRT$$
Where n, R and T is number of moles, gas constant and temperature respectively
For 4 moles $$CO$$
KE$$=\dfrac{3}{2}\times{4}RT$$
For 5 Mole of $${N}_{2}$$
KE$$=\dfrac{3}{2}\times{5}RT$$
For 2 mole of $${C}_{2}{H}_{4}$$
KE$$=\dfrac{3}{2}\times{2}RT$$
$$CO:{N}_{2}:{C}_{2}{H}_{4}=4:5:2$$
Question 9
1 / -0

Which of the following expression is true regarding gas laws?
($$w=$$weight, $$m=$$molar mass)

A
$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 }{ w }_{ 2 } }{ { M }_{ 2 }{ w }_{ 1 } }$$

B
$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 2 }{ w }_{ 1 } }{ { M }_{ 1 }{ w }_{ 2 } }$$

C
$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 }{ w }_{ 1 } }{ { M }_{ 2 }{ w }_{ 2 } }$$

D
$$\dfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\dfrac { { M }_{ 1 }{ w }_{ 1 } }{ { M }_{ 2 }{ w }_{ 2 } }$$

Solution

According to ideal gas equation
$$PV=nRT$$
Where, $$P=$$ Pressure
$$V=$$ Volume
$$n=$$ No. of mole
$$R=$$ Gas constant
$$T=$$ Temperature
$$n=\dfrac {w_1}{M_1}$$
Where $$ {w}_{1}=$$weight of the gas
$${m}_{1}=$$Molar mass
Taking $$P,\ V$$ and $$R$$ constant for two gas 1 and 2

$${ T }_{ 1 }=\dfrac { { P }{ \times V\times }{ M }_{ 1 } }{ { w }_{ 1 }\times { R }_{ } } $$

$${ T }_{ 2 }=\dfrac { { P }{ \times V\times }{ M }_{ 2 } }{ { w }_{2 }\times { R }_{ } } $$

$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 }{ w }_{ 2 } }{ { M }_{ 2 }{ w }_{ 1 } } $$
Question 10
1 / -0

A gas at a pressure of $$5.0$$ bar is heated from $$0^{o}$$ to $$546^{o}C$$ and is simultaneously compressed to one-third of its original volume. Hence, final pressure is:

A
$$\dfrac {5}{9}$$bar

B
$$15.0$$bar

C
$$30.0$$bar

D
$$45.0$$bar

Solution

Initial volume of the gas$$={ V }_{ 1 }L$$
Initial Temperature of the gas$$={ T }_{ 1 }=(0+273)=273K$$
Final pressure of the gas$$={ P }_{ 2 }=?$$
Final volume of the gas$$={ V }_{ 2 }L$$
Final Temperature of the gas$$={ T }_{ 2 }=(546+273)K=819K$$
Given condition is compressed to one third of its original volume.
so $${ V }_{ 1 }=3{ V }_{ 2 }$$
We know from gas laws,

$$\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ 819 } =\dfrac { 5\times 3{ V }_{ 2 } }{ 273 } $$

$${ P }_{ 2 }=\dfrac { 5\times 3\times 819 }{ 273 } bar=45bar$$

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 35

Result

States of Matter Test - 35

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SHARING IS CARING

Question 1

$$0.122$$

$$2.46$$

$$1.22$$

$$24.6$$

Solution

Question 2

$$5.77\ mmHg$$

$$16\ mmHg$$

$$35.08\ mmHg$$

between $$8.77$$ and $$15.54\ mmHg$$

Solution

Question 3

$$4.4$$

$$0.44$$

$$3.6$$

$$1.12$$

Solution

Question 4

$$\dfrac{RT}{PM}$$

$$\dfrac{P}{RT}$$

$$\dfrac{M}{V}$$

$$\dfrac{PM}{RT}$$

Solution

Question 5

$$0.03\ L$$

$$3\ L$$

$$0.3\ L$$

$$1.5\ L$$

Solution

Question 6

$$80$$ torr

$$140$$ torr

$$720$$ torr

$$70$$ torr

Solution

Question 7

$$A,B$$

$$B,C$$

$$B,D$$

$$A,C$$

Solution

Question 8

$$4:5:2$$

$$5:5:4$$

$$2:4:5$$

$$1:1:1$$

Solution

Question 9

$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 }{ w }_{ 2 } }{ { M }_{ 2 }{ w }_{ 1 } }$$

$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 2 }{ w }_{ 1 } }{ { M }_{ 1 }{ w }_{ 2 } }$$

$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 }{ w }_{ 1 } }{ { M }_{ 2 }{ w }_{ 2 } }$$

$$\dfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\dfrac { { M }_{ 1 }{ w }_{ 1 } }{ { M }_{ 2 }{ w }_{ 2 } }$$

Solution

Question 10

$$\dfrac {5}{9}$$bar

$$15.0$$bar

$$30.0$$bar

$$45.0$$bar

Solution

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