States of Matter Test

Result

States of Matter Test - 55

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The density of a liquid is 1.2 g/mL. That are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70):

A
$$\frac{1.2}{35} N_A$$

B
$$\left (\frac{1}{35} N_A \right )^2$$

C
$$\frac{1.2}{(35)^2} N_A$$

D
$$1.2 N_A$$

Solution

Volume of one drop $$=\frac{2}{35}\, mL$$

Moles in one drop $$=\frac{2 \times 1.2}{35 \times 70}=\frac{1.2}{(35)^2}\ mol$$

Number of molecules in one drop $$=\frac{1.2}{(35)^2}\times N_A$$
Question 2
1 / -0

The reaction, $$ZnO(s)+CO(g) \rightleftharpoons Zn(g)+{CO}_{2}(g)$$, has an equilibrium constant of $$1$$ atm at $$1500K$$. The equilibrium partial pressure of zinc vapour in a reaction vessel if an equimolar mixture of $$CO$$ and $${CO}_{2}$$ is brought into contact with solid $$ZnO$$ at $$1500K$$ and the equilibrium is achieved at $$1$$ atm is?

A
$$0.68$$ atm

B
$$0.76$$ atm

C
$$0.24$$ atm

D
$$0.5$$ atm

Solution
Question 3
1 / -0

When a liquid that is immiscible with water was steam distilled at $$95.2^o$$C at a total pressure of $$99.652$$ kPa, the distillate contained $$1.27$$ g of the liquid per gram of water. What will be the molar mass of the liquid if the vapour pressure of water is $$85.140$$ kPa at $$95.2^o$$C?

A
$$99.65\ g mol^{-1}$$

B
$$18 \ g mol^{-1}$$

C
$$134.1 \ g mol^{-1}$$

D
$$105.74\ g mol^{-1}$$

Solution

Total pressure, $$P_{total}=99.652$$ kPa

$$P_{water}=p_B= 85.140$$ kPa

$$P_{liquid}=p_A=(99.652-85.140)$$ kPa

$$=14.512$$ kPa

and $$\displaystyle\dfrac{m_A}{m_B}=\dfrac{1.27}{1}$$

or $$\displaystyle\dfrac{m_A}{m_B}=\dfrac{p_AM_A}{p_BM_B}$$

or, $$M_A=\left(\dfrac{m_A}{m_B}\right)\left(\dfrac{p_BM_B}{p_A}\right)$$

$$M_A=1.27\times \left(\displaystyle \frac{85.140kPa\times 18\ g \ mol^{-1}}{14.512\ kPa}\right)$$

$$M_A\simeq 134.1 \ g \ mol^{-1}$$
Question 4
1 / -0

If one mole monoatomic ideal gas was taken through process $$AB$$ as shown in figure, then select correct option(s).

A
$$w_{AB} = +1496.52\ J$$

B
$$q_{AB} = 5237.82\ J$$

C
$$\triangle H_{AB} = 3741.3\ J$$

D
$$\triangle S_{AB}$$ is $$+ve$$

Solution

Given, $$n = 1\space mole$$ $$PV = nRT$$
$$w_{AB } = P(\Delta V) = P(5) = 5P\space J$$
$$P = - \dfrac{8.314(600)}{15} = 8.314\times 40$$
$$\Rightarrow w_{AB } = - 8.314\times 40 \times 5 \space J$$
Hence, $$w_{AB } =- 1496.52\space J$$
Question 5
1 / -0

An ideal gas undergoes isothermal expansion from ($$10$$ atm, $$1$$L) to ($$1$$ atm, $$10$$L) either by path-I(infinite stage expansion) or by path-II(first against $$5$$atm and then against $$1$$ atm). The value of $$\left(\displaystyle\frac{q_{path-1}}{q_{path-II}}\right)$$ is?

A
$$\displaystyle\frac{2.303}{1.3}$$

B
$$\displaystyle\frac{1.3}{2.303}$$

C
$$\displaystyle\frac{1.0}{13\times 2.303}$$

D
$$\displaystyle 13\times 2.303$$

Solution

For path-I

$$W=-nRT\ln\dfrac { { V }_{ 2 } }{ { V }_{ 1 } } $$$$=-nRT\ln\left( \dfrac { 10 }{ 1 } \right)$$

$$ =-nRT\times 2.303$$

$${ q }_{ 1 }=-W=nRT\times 2.303$$

For Path-II

$$W=-5atm\left( \dfrac { nRT }{ { P }_{ 2 } } -\dfrac { nRT }{ { P }_{ 1 } } \right) -1atm\left( \dfrac { nRT }{ { P }_{ 2 }^{ 1 } } -\dfrac { nRT }{ { P }_{ 1 }^{ 1 } } \right)$$

$$ =-nRT \left( \dfrac { 5 }{ { P }_{ 2 } } -\dfrac { 5 }{ { P }_{ 1 } } \right) + \left( \dfrac { 1 }{ { P }_{ 2 }^{ 1 } } -\dfrac { 1 }{ { P }_{ 1 }^{ 1 }}\right)$$

$$ =-1.3RT.n$$

$$\therefore \quad { q }_{ II }=1.3nRT$$

$$\therefore \quad \dfrac { { q }_{ I } }{ { q }_{ II } } =\dfrac { 2.303 }{ 1.3 }$$

Hence, the correct option is $$\text{A}$$
Question 6
1 / -0

In which of the following case(s) pressure of gas is more than atmospheric pressure.

A

B

C

D

Solution
Question 7
1 / -0

$$18$$ g glucose ($$C_6H_{12}O_6$$) is added to $$178.2$$ g of water. The vapour pressure of this aqueous solution at $$100^0C$$ in torr is:

A
$$7.60$$

B
$$76.00$$

C
$$752.40$$

D
$$759.00$$

Solution

Molecular mass of water $$=\left( 2\times 1 \right) +\left( 1\times 16 \right) =18g$$

For $$178.2$$g water, $${ n }_{ A }=\dfrac{178.2}{18}=9.9\ mol$$

Molecular mass of glucose $$=(6)(12)+(12)(1)+6(16)=180$$g.

For $$18$$g glucose, $${ n }_{ B }=\dfrac{18}{180}=0.1\ mol$$

$${ X }_{ B }=\dfrac{0.1}{\left( 0.1+9.9 \right) }=0.01$$

$${ X }_{ A }=1-0.01=0.99$$

For lowering of vapour pressure,

$$P={ P }_{ A }^{ 0 }{ X }^{ A }={ P }_{ A }^{ 0 }\left( 1-{ X }_{ B } \right) $$

$$P=760\left( 1-0.01 \right) $$

$$P=760-7.6$$

$$P=752.40$$ torr

Vapour pressure of water is $$752.40$$ torr.

Therefore, the correct option is $$C$$.
Question 8
1 / -0

Two liquids $$X$$ and $$Y$$ from an ideal solution. At $$300K$$, a vapour pressure of the solution containing $$1$$ mol of $$X$$ and $$3$$ mol of $$Y$$ is $$550$$ $$mm \ Hg$$. At the same temperature, if $$1$$ mol of $$Y$$ is further added to this solution, a vapour pressure of the solution increased by $$10$$ $$mm \ Hg$$. Vapour pressure ( in mmHg) of $$X$$ and $$Y$$ in their pure states will be respectively:

A
$$300$$ and $$400$$

B
$$400$$ and $$600$$

C
$$500$$ and $$600$$

D
$$200$$ and $$300$$

Solution

$$\begin{array}{l} { P_{ Total } }={ X_{ x } }{ P_{ x } }+{ X_{ y } }{ P_{ y } }\, \, So,\, \, 550=\frac { 1 }{ 4 } { P_{ x } }+\frac { 3 }{ 4 } { P_{ y } } & \left[ \begin{array}{l} Here\, \, { p_{ y } }\, \, and\, \, { p_{ y } }\, \, and\, \, vapour\, \, pressure \\ of\, \, X\, and\, \, Y\, \, in\, \, pure\, \, states. \end{array} \right] \\ { P_{ x } }+3{ P_{ 4 } }=550\times 4;{ P_{ x } }+3{ P_{ y } }=2200\to \left( i \right) & \\ { 2^{ nd } }\, \, case & \\ { P_{ total } }=560\, \, mm\, \, Hg. & \\ \frac { 1 }{ 5 } { P_{ x } }+\frac { 4 }{ 5 } { P_{ y } }=560\, \, { P_{ x } }+{ 48_{ y } }=2800\to \left( { ii } \right) & \\ \left( { ii } \right) -\left( i \right) & \\ { p_{ y } }=600\, \, mm\, \, Hg\, \, { p_{ x } }=2200-1800\Rightarrow 400\, \, mm\, \, 07ng & \end{array}$$
Question 9
1 / -0

At certain temperature ($$T$$) for the gas phase reaction
$$2{ H }_{ 2 }O(g)+2{ Cl }_{ 2 }(g)\rightleftharpoons 4HCl(g)+{ O }_{ 2 }(g);{ K }_{ P }=12\times { 10 }^{ 8 }atm$$
If $${Cl}_{2}$$, $$HCl$$ and $${O}_{2}$$ are mixed in such a manner that the partial pressure of each is $$2atm$$ and the mixture is brough into contact with excess of liquid water. What would be approximate partial pressure of $${Cl}_{2}$$ when equilibrium is attained at temperature ($$T$$)?
[Given:Vapour pressure of water is $$380mm$$ $$Hg$$ at temperature ($$T$$)]

A
$$3.6\times { 10 }^{ -5 }atm$$

B
$${ 10 }^{ -4 }atm$$

C
$$3.6\times { 10 }^{ -3 }atm$$

D
$$0.01$$atm

Solution

$$2H_2O + 2Cl_2 \rightleftharpoons 4HCl + O_2$$
$$K_P = \dfrac{[HCl]^4[O_2]}{[H_2O]^2[Cl_2]^2}$$
Given, partial pressure is $$2\space atm$$ for $$Cl_2, HCl, O_2.$$
$$\Rightarrow 12\times 10^8 = \dfrac{(2)^4(2)}{(2)^2[Cl]^2}$$
Hence, $$P_{Cl_2}$$ is $$3.6\times 10^{-5}\space atm$$
Question 10
1 / -0

$$CuSO_4$$$$\cdot$$$$5H_2O(s)$$$$\rightleftharpoons$$ $$CuSO_4\cdot$$ 3$$H_2O(s)$$ + $$2H_2O(g)$$; $$K_p$$ = 4$$\times10^{-4}$$ $$atm^2$$ if the vapour pressure of water is 38 torr then percentage of relative humidity is: ( Assume all data at constant temperature)

A
$$4$$

B
$$10$$

C
$$40$$

D
$$none\ of\ these$$

Solution

$${K_p} = P_{{H_2}O}^2 = 4 \times {10^{ - 4}}$$
$${P_{{H_2}O}} = 2 \times {10^{ - 2}}\,\,atm \Rightarrow 0.02\,\,atm$$
$$P_{{H_2}O}^1 = 38\,\,atm \Rightarrow \frac{{38}}{{760}}\,\,atm \Rightarrow \frac{1}{{20}}\,\,atm \Rightarrow 0.05\,\,atm$$
$$\% \,\,{\text{Relative}}\,\,humidity = \frac{{{P_{{H_2}O}}}}{{P_{{H_2}O}^1}} \Rightarrow \frac{{0.02}}{{0.05}} = \frac{2}{5} \times 100 \Rightarrow 40\,\,\% $$
Option C is correct.