States of Matter Test

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States of Matter Test - 62

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Weekly Quiz Competition

Question 1
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For the two compounds, the vapour pressure of (2) at a particular temperature is expected to be:-

A
Higher than (i)

B
Lower than that of (i)

C
Same as that of (i)

D
Can be 'higher or lower depending upon the size of the vessel

Solution

(2) has intramolecular Hydrogen bonding. Due to this, there is less association between the molecules and hence, higher vapour pressure.
(1) has intermolecular hydrogen bonding , leading to more association between the molecules. Due to this, (1) has low vapour pressure than (2).
Question 2
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$$100\space g$$ of liquid $$A$$(molar mass $$140\space g { mol }^{ -1 })$$ was dissolved in $$1000\space g$$ of liquid $$B$$ (molar mass $$180 \space g { mol }^{ -1 })$$. The vapour pressure of pure liquid $$B$$ was found to be $$500 \space torr.$$ Calculate the vapour of pure liquid $$A$$ and its vapour pressure in the solution if the total vapour pressure of the solution is $$475 \space torr$$.

A
$$22\space torr$$

B
$$32\space torr$$

C
$$45 \space torr$$

D
$$36 \space torr$$

Solution

Given:-
$$P_{T}=475$$ Torr, $$P_{B}^{\circ}=500$$ Torr
$$\omega_{A}=100\space g, M_{A}=140\space g ; \omega_{B}=1000\space g, M_{B}=180 g$$

To find:- $$P_{A}^{0}$$ and $$P_{A}$$

$$n_{A}=\dfrac{10 0 }{140}=\dfrac{5}{7}, n_{B}=\dfrac{1000}{180}=\dfrac{50}{9}$$

Now, $$x_{A}=\dfrac{n_{A}}{h_{A}+{h_{B}}}=\dfrac{\dfrac{5}{7}}{\dfrac{5}{7}+\dfrac{50}{9}}=0.114$$

$$x_{B}=1-u_{A}=1-0.114=0.886$$

we know, that,

$$\quad P_{T}=P_{A}^{0} x_{A}+P_{B}^{0} x_{B}$$

where $$p_{A} x_{A}=P_{A}$$ and $$p_{B} x_{B}=P_{B}$$

Now, $$\quad P^{0} A \times 0.114+500 \times 0.886=475$$

$$\Rightarrow p^{0} A=280 \cdot 70$$ Torr

$$\therefore \quad P_{A}=p_{A} x_{A}=280.7 \times 0.114$$

So, $$P_A=31.9998$$ Torr
or

$$P_{A} \approx 32$$ Torr
option $$B$$ is correct
Question 3
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In a gaseous mixture at $${20}^{o}C$$ the partial pressure of the components are
$${H}_{2}:150$$ Torr $${CH}_{4}:300$$ Torr
$${CO}_{2}:200$$ Torr $${C}_{2}{H}_{4}:100$$ Torr
Volume percent of $${H}_{2}$$ is:

A
$$26.67$$

B
$$73.33$$

C
$$80.00$$

D
$$20$$

Solution

In a mixture of gases the percent of each gas in the volume is same as the percent of each partial pressure in the total pressure.
Total partial pressure=$$150+300+200+100=750\\ $$torr
$$volume\quad percent H2=\frac { 150 }{ 750 } \times 100\\ =20$$
Question 4
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A $$1.0\ g$$ sample of air consists of approximately $$0.76\ g$$ of nitrogen and $$0.24\ g$$ of oxygen. This sample occupies a $$1.0\ L$$ vessel at $$20^{o}C$$. Then:

A
The partial pressure of $$N_{2}$$ is $$0.65\ atm$$

B
The partial pressure of $$O_{2}$$ is $$0.36\ atm$$

C
The total pressure is $$0.83\ atm$$

D
The total pressure is $$1.05\ atm$$

Solution

conclusion: hence the option (C) is correct.
The data needed for use are
$${ P }_{ { N }_{ 2 }= }?\quad \quad \quad \quad { n }_{ N_{ 2 } }=?\quad \quad \quad V=1.00L\quad \quad \quad T=293K\quad ({ 20 }^{ 0 }C)$$

$${ P }_{ O_{ 2 } }=?\quad \quad \quad { n }_{ { O }_{ 2 } }=?\quad \quad \quad V=1.00L\quad \quad \quad \quad T=293K\quad ({ 20 }^{ 0 }C)$$

The number of $${ n }_{ N_{ 2 } }$$ and $${ n }_{ { O }_{ 2 } }$$ can be obtained from the masses and molar masses (28.02g/mol for $${ N }_{ 2 }$$ 32.00g/mol of $${ O }_{ 2 }$$0

Moles of $${ N }_{ 2 }$$= 0.76G $${ N }_{ 2 }$$x =$$\dfrac { 1\quad mol\quad { N }_{ 2 } }{ 28.02g\quad { N }_{ 2 } } =0.0271\quad mol\quad { N }_{ 2 }$$

Moles of $${ O }_{ 2 }$$=0.24g$${ O }_{ 2 }$$x=$$\dfrac { 1\quad mol\quad { O }_{ 2 } }{ 32.00{ gO }_{ 2 } } =0.00750mol\quad { O }_{ 2 }$$
using $${ P }_{ A }=\dfrac { { n }_{ A }RT }{ V } $$
$${ P }_{ { N }_{ 2 }= }0.0271mol{ N }_{ 2 }\times \dfrac { 0.08206Latm }{ 1Kmol } \times \dfrac { 293 }{ 1.00L } =0.65\quad atm{ N }_{ 2 }$$

$${ P }_{ O_{ 2 } }=0.00750mol{ O }_{ 2 }\times \dfrac { 0.08206Latm }{ 1Kmol } \times \dfrac { 293 }{ 1.00L } =0.18\quad atm{ O }_{ 2 }$$

The total pressure is the sum of these partial pressure

$$P=0.65 atm+0.18atm=0.83atm$$
Question 5
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The K.E of $$N$$ molecule of $$O_{2}$$ is $$xjoules$$ at__ $${123}^{o}C$$. Another sample of $$O_{2}$$ at $${327}^{o}C$$ has a K.E of $$2x$$ joules. The latter sample contains:

A
$$N$$ molecules of $$O_{2}$$

B
$$2N$$ molecules of $$O_{2}$$

C
$$N/2$$ molecules of $$O_{2}$$

D
$$N/4$$ molecules of $$O_{2}$$

Solution

Though the individual speeds are changing, the distribution of speeds remains constant at a particular temperature.
$$K.E_{ O_{ 2 } }=\dfrac { \dfrac { 3 }{ 2 } \times \dfrac { N }{ 32 } \times R\times 150 }{ \dfrac { 3 }{ 2 } \times \dfrac { { N }' }{ 32 } \times R\times 300 } $$
             $$=\dfrac { x }{ 2x } $$
     $$K.E_{ O_{ 2 } }=\dfrac { N\times 1 }{ N'\times 2 } $$

              $$N=N'$$

conclusion: hence the option (A) is correct.
Question 6
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$$0.5\ \text{mole}$$ of each $$H_{2},SO_{2}$$ and $$CH_{4}$$ are kept in a container. A hole was made in the container. After 3 hours, decreasing order of partial pressures of gases in the container will be:

A
$$P_{SO_{2}} > P_{CH_{4}} > P_{H_{2}}$$

B
$$P_{H_{2}} > P_{SO_{2}} > P_{CH_{4}}$$

C
$$P_{CH_{4}} > P_{SO_{4}} > P_{H_{2}}$$

D
$$P_{CH_{4}} > P_{H_{2}} > P_{SO_{2}}$$

Solution

Answer : D.  $$P_{SO_{2}} > P_{CH_{4}} > P_{H_{2}}$$

The partial pressure of a gas is directly proportional to mole fraction, therefore directly proportional to the number of moles.

Initially, there are equal number of moles (0.5 moles) of each gases. So their partial pressures are equal.

After 3 hours, some amount of each gases will be effused out through the hole. Gas with larger effusion rate will be effused more through the hole and will have less number of moles remaining in the container, resulting in lower partial pressure compared to others.

We know that,
Molar mass of $$H_{2}$$, $$M_{H_{2}}$$ = $$1\times2$$ = $$2$$ g

Molar mass of $$SO_{2}$$, $$M_{SO_{2}}$$ = $$32 + (16\times2)$$ = $$64 $$ g

Molar mass of $$CH_{4}$$, $$M_{CH_{4}}$$ = $$12 + (1\times4)$$ = $$16$$ g

Therefore, the order of molar masses,
$$M_{SO_{2}} > M_{CH_{4}} > M_{H_{2}}$$

$$\textrm{Rate of effusion of gas}$$ $$\alpha$$ $$\dfrac{1}{\sqrt{\textrm{molar mass}}}$$
$$\Rightarrow$$ Order of rates of effusion,
$$r_{SO_{2}} < r_{CH_{4}} < r_{H_{2}}$$

$$\textrm{Number of moles remaining in the container}$$  $$\alpha$$ $$\dfrac{1}{\textrm{Rate of effusion}}$$
$$\Rightarrow$$ Order of number of moles remaining in the container,
$$n_{SO_{2}} > n_{CH_{4}} > n_{H_{2}}$$

$$\textrm{Partial pressure}$$  $$\alpha$$ $$\textrm{Number of moles remaining in the container}$$
$$\Rightarrow$$ Order of partial pressures,
$$P_{SO_{2}} > P_{CH_{4}} > P_{H_{2}}$$
Question 7
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$$N_{2}$$ is found in a litre flask under $$100\ kPa$$ pressure and $$O_{2}$$ is found in another $$3$$ litre flask under $$320\ kPa$$ pressure. If the two flask are connected, the resultant pressure is:

A
$$265\ kPa$$

B
$$210\ kPa$$

C
$$420\ kPa$$

D
$$365\ kPa$$

Solution

At initial condition $${ P }_{ 1 }=100kPa$$ $$=1atm{ V }_{ 1 }$$
$$=1L{ P }_{ 2 }$$
$$=320k{ Pa=3atm{ V }_{ 2 } }$$=$$3L$$
Applying ideal gas law,$$\dfrac { PV }{ RT } =n$$
total mole at initial conditions,
$$(1+{ n }_{ 2 })$$=$${ n }_{ i }=\dfrac { 2 }{ RT } +\dfrac{ 9.3 }{ RT } =\dfrac { 11.3 }{ RT } $$
After mixing the gases the number of moles remains constant gases do not undergo reaction.
$${ n }_{ i }={ n }_{ f }={ P }_{ f }\times \dfrac { { V }_{ f } }{ RT } \\ Where,{ P }_{ f }=final\quad pressure\\ \quad \quad \quad \quad \quad { V }_{ f }=final\quad volume\\ { V }_{ f }={ V }_{ 1 }+{ V }_{ 2 }=4L\\ thus,{ P }_{ f }\times \dfrac { 4 }{ RT } =\dfrac { 11.3 }{ RT } \\ { P }_{ f }=2.85atm$$
$${ P }_{ f }=285kPa\approx 265kPa$$
Question 8
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The density of a gas $$A$$ is twice that of $$B$$. Molar mass of $$A$$ is half that of $$B$$. The ratio of partial pressure of $$A$$ to $$B$$ is:

A
$$1/4$$

B
$$1/2$$

C
$$4/1$$

D
$$2/1$$

Solution

According to Ideal gas equation
$$PV=nRT$$

$$n=\dfrac{m}{M}$$

$$PV=\dfrac { m }{ M } RT\\ P=\dfrac { m }{ V } \times \dfrac { RT }{ M } \\ P=d \times \ \dfrac { RT }{ M } $$

As $$R$$ and $$T$$ are constant

$$ P_{ A }=\dfrac { { d }_{ A } }{ { M }_{ A } } \\ P_{ B }=\dfrac { { d }_{ B } }{ { M }_{ B } } $$

Given $${ M }_{ A }=\dfrac { 1 }{ 2 } { M }_{ B }\\ d_{ A }=2d_{ B }$$

$$\dfrac { { P }_{ A } }{ { P }_{ B } } =\dfrac { { d }_{ A }{ M }_{ B } }{ { M }_{ A }{ d }_{ B } } $$

Putting the value of $${ d }_{ A }$$ and $${ M }_{ A }$$ in the above equation
$$\dfrac { { P }_{ A } }{ { P }_{ B } } =4$$
Question 9
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A container is divided into two compartment. One compartment contains $$2$$ mole of $$N_{2}$$ gas at $$1$$ atm and $$300\ K$$ & other compartment contains $$H_{2}$$ gas at the same temperature and pressure. Volume of $$H_{2}$$ compartment is four times the volume of $$N_{2}$$ compartment. [assume no reaction under these condition]
If the container containing $$N_{2}$$ and $$H_{2}$$ are further heated to $$1000\ K$$ forming $$NH_{3}$$ with $$100\%$$ yield. Calculate the final total pressure.

A
$$2.22\ atm$$

B
$$3\ atm$$

C
$$2\ atm$$

D
$$3.33\ atm$$

Solution

Compartment containing $$N_2$$
$$n=2$$
$$P=1$$atm
$$T=300$$K
Finding for V
Using Ideal gas equation
$$PV=nRT$$
$$V=\dfrac{2\times0.0821\times300}{1}=49.26$$L
Compartment containing $$H_2$$
Givn volume is 4 times than $$N_2$$
$$V_2 =4\times49.26=197$$L
Finding moles for $$H_2$$ Compartment
given pressure and temperature is equal to $$N_2$$ compartment
$$n=\dfrac{197}{0.0821\times300}=8$$
Calculating final pressure
$$T=1000$$K
$$P=\dfrac{8\times0.0821\times1000}{197}=3.33$$atm
Question 10
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A glass tube of volume $$112ml$$. containing a gas is partially evacuated till the pressure in it drops to $$3.8 \times 10^{-5}\ torr$$ at $$0^{o}C$$. The number of molecules of the gas remaining in the tube is?

A
$$3\times 10^{17}$$

B
$$1.5\times 10^{14}$$

C
$$4.5\times 10^{18}$$

D
$$6\times10^{18}$$

Solution

Pressure $$=3.8\times 10^{-5} torr$$ and temperature $$=0^oc+273k$$

$$R=62.36367\ L\ Torr |c^{-1} mo|^{-1}$$

Volume $$=112\ ml =112 \times 10^{-3}c$$

number of mole $$n=\dfrac{PV}{RT}$$

$$=\dfrac{3.8\times 10^{-5}\times 112\times10^{-3}}{62.36367\times 273}$$

$$=2.5\times10^{-10}$$

number of molecules of gas $$=6.022\times 10^{23} \times 2.5\times 10^{-10}$$

$$=1.5\times10^{14}$$

So, the correct option is $$B$$

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States of Matter Test - 62

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States of Matter Test - 62

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SHARING IS CARING

Question 1

Higher than (i)

Lower than that of (i)

Same as that of (i)

Can be 'higher or lower depending upon the size of the vessel

Solution

Question 2

$$22\space torr$$

$$32\space torr$$

$$45 \space torr$$

$$36 \space torr$$

Solution

Question 3

$$26.67$$

$$73.33$$

$$80.00$$

$$20$$

Solution

Question 4

The partial pressure of $$N_{2}$$ is $$0.65\ atm$$

The partial pressure of $$O_{2}$$ is $$0.36\ atm$$

The total pressure is $$0.83\ atm$$

The total pressure is $$1.05\ atm$$

Solution

Question 5

$$N$$ molecules of $$O_{2}$$

$$2N$$ molecules of $$O_{2}$$

$$N/2$$ molecules of $$O_{2}$$

$$N/4$$ molecules of $$O_{2}$$

Solution

Question 6

$$P_{SO_{2}} > P_{CH_{4}} > P_{H_{2}}$$

$$P_{H_{2}} > P_{SO_{2}} > P_{CH_{4}}$$

$$P_{CH_{4}} > P_{SO_{4}} > P_{H_{2}}$$

$$P_{CH_{4}} > P_{H_{2}} > P_{SO_{2}}$$

Solution

Question 7

$$265\ kPa$$

$$210\ kPa$$

$$420\ kPa$$

$$365\ kPa$$

Solution

Question 8

$$1/4$$

$$1/2$$

$$4/1$$

$$2/1$$

Solution

Question 9

$$2.22\ atm$$

$$3\ atm$$

$$2\ atm$$

$$3.33\ atm$$

Solution

Question 10

$$3\times 10^{17}$$

$$1.5\times 10^{14}$$

$$4.5\times 10^{18}$$

$$6\times10^{18}$$

Solution

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