States of Matter Test

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States of Matter Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Equal masses of $$SO_2, CH_4$$ and $$O_2$$ are mixed in empty container at $$298$$K, where total pressure is $$2.1$$ atm. The partial pressures of $$CH_4$$ in the mixture is:

A
$$0.5$$ atm

B
$$0.75$$ atm

C
$$1.2$$ atm

D
$$0.6$$ atm

Solution

Partial pressure is directly proportional to the mole fraction of the gas
Partial pressure $$=$$ Total pressure $$\times$$ Mole fraction

Let the mass of each constituent $$= x$$
Moles of $$SO_2$$ $$=$$ $$\frac{x}{64}$$

Moles of $$CH_4$$ $$=$$ $$\frac{x}{16}$$

Moles of $$O_2$$ $$=$$  $$\frac{x}{32}$$

Total Moles $$=$$  $$\frac{7x}{64}$$

Partial pressure $$=mole\ fraction \times total\ pressure =$$  $$\frac{\frac{x}{16}}{\frac{7x}{64}}$$ x21

$$= 1.2\ atm$$

Hence partial pressure of $$CH_4= 1.2\ atm$$
Question 2
1 / -0

The vapour pressure of a pure liquid A is $$70$$ torr at $$27^o$$C. 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 $$27^o$$C. The vapour pressure of pure liquid B at $$27^o$$C.

A
$$14$$

B
$$56$$

C
$$140$$

D
$$70$$

Solution

Here we will use the simple total vapour pressure equation:

$$P_T=x_AP_A^o+x_BP_B^o$$

$$\Rightarrow 84\ torr= 0.8 \times 70\ torr+ 0.2 \times P_B^o$$

$$\Rightarrow P_B^o=140\ torr$$ .
Question 3
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If 2 moles of A and 3 moles of B are mixed to form an ideal solution vapour pressure of A and B are 120 and 180 mm of Hg respectively. then the composition of A and B in the Vapour phase when the first traces of vapour are formed in the above case is:

A
$$X^l A = 0.407$$

B
$$X^l A = 0.8$$

C
$$X^l A = 0.109$$

D
$$X^l A = 0.307$$

Solution

$$X_A=\dfrac{2}{2+3}=\dfrac{2}{5}$$
$$X_B=\dfrac{3}{2+3}=\dfrac{3}{5}$$
$$X^1_A=\dfrac{P^0_A\times A}{P_{total}}$$
$$=\dfrac{120\times \dfrac{2}{5}}{120\times \dfrac{2}{5}+180\times \dfrac{3}{5}}$$
$$=\dfrac{48}{48+108}$$
$$X^1_A=0.307$$
Question 4
1 / -0

The mass of $$350{ cm }^{ 3 }$$ of a diatomic gas at 273 K at 2 atmospheres pressure is one gram. Calculate the mass of one atom of the gas.

A
$$5.3\times 10^{-23}g$$

B
$$2.65\times 10^{-23} g$$

C
$$12.02\times 10^{23}g$$

D
None of these

Solution

$$n=\dfrac{PV}{RT}$$
$$=\dfrac{2\times 0.35}{0.082\times 273}$$
$$\approx 0.031 mol$$
$$0.031$$ mol weight $$1g$$
$$0.031\times 6022\times 10^{23}atoms$$= $$1g$$
$$=\dfrac{1}{0.031\times 6.022}\times 10^{-23}$$
$$=5.3\times 10^{-23}g$$
Question 5
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Starting with one mole of nitrogen and 3 moles of hydrogen, at equilibrium $$50\%$$ of each had reacted according to the reaction: $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$
If the equilibrium pressure is $$P$$, the partial pressure of hydrogen at equilibrium would be:

A
$$\dfrac{P}{2}$$

B
$$\dfrac{P}{3}$$

C
$$\dfrac{P}{4}$$

D
$$\dfrac{P}{6}$$

Solution

$${ N }_{ 2 }+{ 3H }_{ 2 }\longrightarrow 2{ NH }_{ 3 }$$
$$x$$ $$3x$$ $$0$$ $$\longrightarrow $$ Initial
$$\left( x-\dfrac { x }{ 2 } \right) \left( 3x-\dfrac { 3x }{ 2 } \right) \left( \dfrac { 2x }{ 2 } \right) \longrightarrow $$ Eqn ($$50$$% of $${ N }_{ 2 },{ H }_{ 2 }$$ has reacted)

Total moles of eqn $$=x-\dfrac { x }{ 2 } +3x-\dfrac { 3x }{ 2 } +x=3x$$

Now, $${ P }_{ { H }_{ 2 } }=\dfrac { Moles\ of\ { H }_{ 2 } }{ Total\ moles } \times $$ Total pressure

$$=\left( \dfrac { 3x-3x/2 }{ 3x } \right) \times P=\dfrac { P }{ 2 } $$
Question 6
1 / -0

The Total pressure in a mixture of $$8g$$ of dioxygen and $$4g$$ of dihydrogen confined in a vessel of $$1dm^34$$ at $$27^oC$$ is ?

A
$$R = 83bar\ dm^3k^{-1}mol^{-1}$$

B
$$R = 93bar\ dm^3k^{-1}mol^{-1}$$

C
$$R = 56bar\ dm^3k^{-1}mol^{-1}$$

D
$$R = 63bar\ dm^3k^{-1}mol^{-1}$$

Solution

Mass of oxygen= $$8g= 8/32=0.25 mol$$
Mass of hydrogen= $$4g=4/2=2 mol$$
From $$PV=nRT$$
$$\Rightarrow P \times 1=(0.25+2)0.083 \times 300= 56.02$$ bar
Question 7
1 / -0

Calculate the volume occupied by 16 gram $$O_2$$ at 300K and 8.31 MPa if $$\dfrac{P_cV_c}{RT_c}$$ = 3/8 and $$\dfrac{P_rV_r}{T_r}$$=2.21.
(Given : R = 8.314 J/ K-mol)

A
125.31 mL

B
124.31 mL

C
248.62 mL

D
None of these
Question 8
1 / -0

A Solution of non-volatile solute in water freezes at $$-0.{ 30 }^{ }C.$$ The vapour-pressure of pure water at $$298K$$ is $$23.51mm Hg$$ and $${ K }_{ 1 }$$ for water is $$1.86 K. kg { mol }^{ -1 }$$ Calculate the vapour-pressure of this solution at $$298K.$$

A
$$23.4mm$$

B
$$24.8mm$$

C
$$34.8mm$$

D
$$40mm$$

Solution

We know that, $$\triangle { T }_{ f }={ K }_{ f }m$$
$$\therefore \quad m=\cfrac { \triangle { T }_{ f } }{ { K }_{ f } } =\cfrac { 0.30 }{ 1.86 } =0.161$$
According to Raooh's law,
$${ P }^{ \circ }-P/{ P }^{ \circ }=$$No. of moles of Solute/No of moles of solvent
$$\therefore \quad 23.51-P/23.51=\cfrac { 0.161 }{ 1000/18 } =0.0020898$$
$$\therefore \quad P=23.51-23.51\times 0.0020898$$
$$\therefore \quad P=23.44mmHg$$
Question 9
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In the following graphical representation for reaction $$A \to B$$ there are two types of regions:

A
$$I$$ and $$II$$ both represent kinetic region at different interval

B
$$I$$ and $$II$$ both represent equilibrium regions at different time interval

C
$$I$$ represents kinetic while $$II$$ represents equilibrium region

D
$$I$$ represents equilibrium while $$II$$ represents kinetic region

Solution

I represents kinetic region and II represents equilibrium region.
The correct option is C.
Question 10
1 / -0

Which of the following gases has the highest density under standard conditions?

A
$$CO$$

B
$$N_2O$$

C
$$C_3H_8$$

D
$$SO_2$$

Solution

Under Standard conditions The Compound with Highest Molecular has Highest Density (PM = $$\rho$$RT)

$$S{O}_{2}$$ has high molecular weight so it has the highest density

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 33

Result

States of Matter Test - 33

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Question 1

$$0.5$$ atm

$$0.75$$ atm

$$1.2$$ atm

$$0.6$$ atm

Solution

Question 2

$$14$$

$$56$$

$$140$$

$$70$$

Solution

Question 3

$$X^l A = 0.407$$

$$X^l A = 0.8$$

$$X^l A = 0.109$$

$$X^l A = 0.307$$

Solution

Question 4

$$5.3\times 10^{-23}g$$

$$2.65\times 10^{-23} g$$

$$12.02\times 10^{23}g$$

None of these

Solution

Question 5

$$\dfrac{P}{2}$$

$$\dfrac{P}{3}$$

$$\dfrac{P}{4}$$

$$\dfrac{P}{6}$$

Solution

Question 6

$$R = 83bar\ dm^3k^{-1}mol^{-1}$$

$$R = 93bar\ dm^3k^{-1}mol^{-1}$$

$$R = 56bar\ dm^3k^{-1}mol^{-1}$$

$$R = 63bar\ dm^3k^{-1}mol^{-1}$$

Solution

Question 7

125.31 mL

124.31 mL

248.62 mL

None of these

Question 8

$$23.4mm$$

$$24.8mm$$

$$34.8mm$$

$$40mm$$

Solution

Question 9

$$I$$ and $$II$$ both represent kinetic region at different interval

$$I$$ and $$II$$ both represent equilibrium regions at different time interval

$$I$$ represents kinetic while $$II$$ represents equilibrium region

$$I$$ represents equilibrium while $$II$$ represents kinetic region

Solution

Question 10

$$CO$$

$$N_2O$$

$$C_3H_8$$

$$SO_2$$

Solution

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