States of Matter Test

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States of Matter Test - 56

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Weekly Quiz Competition

Question 1
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A vessel has $$6 \ g$$ of oxygen at pressure $$P$$ and temperature $$400 \ K$$. A small hole is made in it so that oxygen $$O_2$$ leaks out. How much oxygen leaks out if the final pressure is $$P/2$$ and temperature $$300 \ K$$?

A
$$5 \ g$$

B
$$4 \ g$$

C
$$2 \ g$$

D
$$3 \ g$$

Solution

By ideal gas equation,
$$PV=nRT$$

$$\therefore P_{1}V=n_{1}RT_{1}$$

$$ P_{2}V=n_{2}RT_{2}$$

$$P_2 = \dfrac{P_1}{2}$$

$$\cfrac{n_{1}}{n_{2}}=\cfrac{P_{1} \times T_{2}}{P_{2}\times T_{1}}=\cfrac{P_{1}\times 300}{P_{2} \times 400}$$

$$\cfrac{n_{1}}{n_{2}}=1.5$$

Now, $$n_{1}=\cfrac{mass}{molar\,\, mass}=\cfrac{6}{32}$$

$$n_{2}=\cfrac{m}{32}$$

$$\therefore \cfrac{n_{1}}{n_{2}}=1.5=\cfrac{\cfrac{6}{32}}{\cfrac{m}{32}}$$

$$m=\cfrac{6}{1.5}=4\, gm$$.

$$\therefore$$ Mass of oxygen left$$=4 \, gm$$

$$\therefore $$ Mass of oxygen leaked $$=6-4=2\,gm$$
Question 2
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Dalton's law of partial pressure will not apply to which of the following mixture of gases?

A
$$H_{2}$$ and $$SO_{2}$$

B
$$H_{2}$$ and $$Cl_{2}$$

C
$$H_{2}$$ and $$CO_{2}$$

D
$$CO_{2}$$ and $$Cl_{2}$$
Question 3
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Equal mass of $$H_2$$, $$He$$ and $$CH_4$$ are mixed in empty container at $$300$$K, when total pressure is $$2.6$$ atm. The partial pressure of $$H_2$$ in the mixture is:

A
$$2.1$$ atm.

B
$$1.6$$ atm.

C
$$0.8$$ atm.

D
$$0.2$$ atm.

Solution

According to Dalton's law, the partial pressure of a gas is equals to the product of total pressure and mole fraction.
We have $$H_2,He\:and\:CH_4$$ gas with equal mass, m.
Total Pressure = 2.6 atm
$$Partial\:Pressure of {H_2}=Total\:Pressure\times\chi_{H_2}$$
$$=2.6\times\dfrac{n_{H_2}}{n_{H_2}+{n_{He}}+{n_{CH_4}}}$$

$$=2.6\times\dfrac{\frac{m}{2}}{\frac{m}{2}+\frac{m}{4}+\frac{m}{16}}$$

$$=2.6\times\dfrac{8}{13}$$
$$Partial\:Pressure_{H_2}=1.6\:atm$$
Question 4
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A gaseous mixture contains $$56\ g\ N_{2}. 44\ g\ CO_{2}$$ and $$16\ g\ CH_{4}$$. The total pressure of the mixture is $$720\ mm\ Hg$$. The partial pressure of $$CH_4$$ in mm Hg is:

A
$$620\ mm$$ of $$Hg$$

B
$$180\ mm$$ of $$Hg$$

C
$$160\ mm$$ of $$Hg$$

D
$$200\ mm$$ of $$Hg$$

Solution

Given: Mass of $$N_{2}$$ = 56g

             Mass of $$CO_{2}$$ = 44g

             Mass of $$CH_{4}$$ = 16g

Solution: Moles of $$N_{2}$$ = $$\dfrac{56}{28}$$ = 2

                 Moles of $$CO_{2}$$ = $$\dfrac{44}{44}$$ = 1

                 Moles of $$CH_{4}$$ = $$\dfrac{16}{16}$$ = 1

Now, as we know, partial pressure of a gaseous component = mole fraction of the component * total pressure

Thus, partial pressure of $$CH_{4}$$ = ($$\dfrac{1}{2 + 1 +1}$$ * 720) mm Hg


= 180 mm Hg

Thus, the correct option is (B).
Question 5
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The vapour pressure of pure benzene at a certain temperature is $$640\ mm\ Hg$$. A non-volatile solute weighing $$2.175\ g$$ is added to $$39.0\ g$$ of benzene. The vapour pressure of solution is $$600\ mm\ Hg$$. What is the molar mass of the solute?

A
$$72.1\ g\ mol^{-1}$$.

B
$$70.6\ g\ mol^{-1}$$.

C
$$69.4\ g\ mol^{-1}$$.

D
$$56.2\ g\ mol^{-1}$$.

Solution

$$ \begin{aligned} P_{A}^{0} &=640 \mathrm{~mm}{\mathrm{Hg}} \\ P_{S} &=600 \mathrm{~mm} {\mathrm{Hg}} \\ & \omega_{A}=39 \mathrm{~g} ; \quad \omega_{B}=2.175 \mathrm{~g} \end{aligned} $$
$$ \begin{array}{l}\text{According to Rault's law (solution }\\ \text { is ideal or dilute), } \frac{P_{A}^{\circ}-P_{S}}{P_{S}}=\frac{n_{B}}{n_{A}}=\frac{\omega_{B}}{M_{B}} \times \frac{M_{n}}{\omega_{A}}\end{array} $$
$$ \begin{array}{l} \Rightarrow \frac{640-600}{600}=\frac{2.175}{M _B} \times \frac{78}{39} \\ \Rightarrow \quad M_{B}=\frac{2.175 \times 78}{39} \times \frac{600}{40}=65.25 \\ \therefore \text { Molar mass of solute }=65.25 \mathrm{~g/mol} \end{array} $$
Question 6
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A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 k Calculate vapour pressure of water at 298K?

A
$$4.53kPa$$

B
$$3.53kPa$$

C
$$5.53kPa$$

D
$$6.53kPa$$
Question 7
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The vapour pressure of $$C_6H_6$$ and $$C_7H_8$$ mixture at $$50^{\circ}C$$ is given by $$p=179X_B+92$$ where $$X_B$$ is the mole fraction of $$C_6H_6$$.
Calculate (in mm) Vapour pressure of liquid mixture obtained by mixing $$936 \, g\ C_6H_6 \, $$ and $$736\ g$$ toluene is:

A
300 mm Hg

B
250 mm Hg

C
199.4 mm Hg

D
180.6 mm Hg

Solution

Given,PM = 179 $$X_B$$+92
For pure $$C_{6} h_{6}$$, $$x_{B},$$ i.e. mole fraction of benzene $$=1$$
$$\Rightarrow \quad P_{B}^{0}=179+92=271\mathrm{~mm}$$
$$x_{B}=\frac{\frac{936}{78}}{\frac{936}{78}+\frac{736}{92}}=\frac{12}{12+8}=0.60$$
$$\begin{aligned} x_{T} &=\frac{\frac{736}{92}}{\frac{736}{92}+\frac{936}{78}}=\frac{8}{12+8}=0.40 \\ P_{M}=& 179 x_{B}+92 \\=& 179 \times 0.6+92 \\=& 107.4+92 \\=& 199.4 \mathrm{~mm} \text { of } \mathrm{Hg} \\ \text { Hence, }(\mathrm{C}) \text { is correct option. } \end{aligned}$$
Question 8
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Pressure of the gas in column (1) is :

A
60 cm of Hg

B
55 cm of Hg

C
50 cm of Hg

D
45 cm of Hg

Solution
Question 9
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Vapour pressure of $$CC{l}_{4}$$ at $$25^{O}C$$ is 143mm Hg. 0.5gm of a non-volatile solute (mol. wt. 65) is dissolved in 100ml of $$CC{l}_{4}$$. Find the vapour pressure of the solution.
(Density of $$CC{l}_{4} = 1.58 gm/{cm}^{3}$$)

A
141.93 mm

B
94.39 mm

C
199.34 mm

D
143.99 mm

Solution

Here logically the option A is correct because when the non-volatile solute is added is added to the solvent the vapour pressure of the solution gets lowered and here only option A has a vapour pressure less than the vapour pressure of the solvent.
Question 10
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A sample of pure $$NO_{2}$$ gas healed lo 1000 K decomposes.

$$NO_{2}(g) \leftrightharpoons 2NO(g) + O_{2}(g)$$

The equillibrium constant $$K_{P}$$ is 100atm. Analysis shows that the partial pressure of $$O_{2}$$ is 0.25 atm at equillibrium. The partial pressure of $$NO_{2}$$ at equillibrium is:

A
0.03

B
0.25

C
0.0245

D
0.04

Solution

$$2NO_{2}(g) \leftrightharpoons 2NO(g) + O_{2}(g)$$
Analysis shows that the partial pressure of $$O_{2}$$ is 0.25 atm at equilibrium. The partial pressure of $$NO$$ at equilibrium $$\displaystyle 2 \times $$ the partial pressure of $$O_{2}$$ $$\displaystyle 2 \times 0.25=0.50$$ atm.
The equillibrium constant $$K_{P}$$ is 100 atm.
$$\displaystyle K_p=\dfrac { P_{ NO}^2 \times P_{ O2} }{ P_{ NO2} ^2} $$
$$\displaystyle 100=\dfrac {(0.50)^2 \times 0.25}{ P_{ NO2} ^2 }$$
$$\displaystyle P_{ NO2} ^2=0.000625$$
$$\displaystyle P_{ NO2} =0.025 \simeq 0.03$$
The partial pressure of $$NO_{2}$$ at equilibrium is 0.03 atm.

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States of Matter Test - 56

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States of Matter Test - 56

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Question 1

$$5 \ g$$

$$4 \ g$$

$$2 \ g$$

$$3 \ g$$

Solution

Question 2

$$H_{2}$$ and $$SO_{2}$$

$$H_{2}$$ and $$Cl_{2}$$

$$H_{2}$$ and $$CO_{2}$$

$$CO_{2}$$ and $$Cl_{2}$$

Question 3

$$2.1$$ atm.

$$1.6$$ atm.

$$0.8$$ atm.

$$0.2$$ atm.

Solution

Question 4

$$620\ mm$$ of $$Hg$$

$$180\ mm$$ of $$Hg$$

$$160\ mm$$ of $$Hg$$

$$200\ mm$$ of $$Hg$$

Solution

Question 5

$$72.1\ g\ mol^{-1}$$.

$$70.6\ g\ mol^{-1}$$.

$$69.4\ g\ mol^{-1}$$.

$$56.2\ g\ mol^{-1}$$.

Solution

Question 6

$$4.53kPa$$

$$3.53kPa$$

$$5.53kPa$$

$$6.53kPa$$

Question 7

300 mm Hg

250 mm Hg

199.4 mm Hg

180.6 mm Hg

Solution

Question 8

60 cm of Hg

55 cm of Hg

50 cm of Hg

45 cm of Hg

Solution

Question 9

141.93 mm

94.39 mm

199.34 mm

143.99 mm

Solution

Question 10

0.03

0.25

0.0245

0.04

Solution

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