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CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 14 - MCQExams.com
CBSE
Class 11 Medical Chemistry
States Of Matter Gases And Liquids
Quiz 14
Two gases X and Y have densities, $${d_{(x)}} = 3{d_{(y)}}$$ and molecular masses, $${M_{(x)}} = 0.5{M_{(y)}}$$. Then, the ratio of their pressures, i.e, $${P_x}:{P_y}$$ would be
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1/4
0%
1/6
0%
4
0%
6
The vapour pressure of water at T(K) is 20 mm Hg. The following solutions are prepared at T(K) :
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6 g of urea (mol. wt. = 60) is dissolved in 178.2 g of water
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0.01 mole of glucose is dissolved in 179.82 g of water
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5.3 g of $$N{a_2}C{O_3}$$ (mol. wt. =106) is dissolved in 179.1 g of water.
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non of these
If $${X_A}$$ and $${X_B}$$ represent mole fraction in liquid phase while $${y_A}$$ and $${y_B}$$ represent mole fraction in vapour phase in binary solution. Which of the following statements is/are correct?
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At point C liquid and vapour phase have same composition
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Vapour pressure of B is less than A
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Intermolecular forces in the solution are weaker than between like particles
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All of these
1.2 L of oxygen at a constant pressure of 2.00 atm was kept in a cylinder and provided 10.00 K cal of heat. The volume of oxygen increases to 1.8 L. The value of $$ \Delta E $$ is
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9.970 Kcal
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0.9970 Kcal
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10.100 Kcal
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10.970 Kcal
Find the slope of the curve plotted between $$\mathrm { P }$$ Vs $$T$$ for closed container of volume $$2$$$$\mathrm { L }$$ having same moles of gas:
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$$\dfrac { e } { 2000 }$$
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$$2000$$$$\mathrm { e }$$
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$$500$$$$\mathrm { e }$$
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$$\dfrac { 2} { 1000 e }$$
Pressure exerted by 2 grams of helium present in a vessel is 1.5 atm. If 4 grams of gas 'X' is introduced into the same vessel keeping the condition constant, the pressure is 2.25 atm. Gas 'x' is
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hydrogen
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methane
0%
oxygen
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soldier oxide
The figure shows the graphs of pressure versus density for an ideal gas at two temperatures, $$T_1$$ and $$T_2$$, according to this which is correct?
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$$T_1 > T_2$$
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$$T_1 = T_2$$
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$$T_1 < T_2$$
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none of the above
An ideal solution contains two volatile liquids $$A (P^0=100 \ torr)$$. If the mixture contains 1 mole of A and 3 moles of B, the total vapour pressure of distillate is:
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150 torr
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188.88 torr
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185.72 torr
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198.88 torr
A bulb of 3 L capacity filled with air is heated from $${\text{2}}{{\text{7}}^{\text{o}}}{\text{C}}\;{\text{to}}\;{{\text{t}}^{\text{o}}}{\text{C}}$$. The air thus, expelled measured 1.45 L at $${\text{1}}{{\text{7}}^{\text{o}}}{\text{C}}$$ Considering the pressure to be 1 atm throughout the experiment and ignoring the experiment of bulb, the value of t is
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300 k
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$$\;{300^{\text{o}}}{\text{C}}$$
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327 k
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$$\;{327^{\text{o}}}{\text{C}}$$
Two liquids A & B form an ideal solution. What is the vapour pressure of solution containing 2 moles of A and 3 moles of B at 300 K? [Given: At 300 K, Vapour pr. of pure liquid A $$\left( P\begin{matrix} 0 \\ A \end{matrix} \right) $$=100 torr, Vapour pr. of pure liquid B $$\left( P\begin{matrix} 0 \\ B \end{matrix} \right) $$=300 torr]
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200 torr
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140 torr
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180 torr
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None of these
For which of the following reactions, Gay Lussac's law is not applicable:
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Formation of $$HI$$ from its constituents
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Formation of $$N H_{3}$$ from its constituents
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Formation of $$C O_{2}$$ from its constituents
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Formation of $$
S O_{3}
$$ from $$
S O_{2}
$$ and $$
O_{2}
$$
The law which suggests $$n_1 = n_2$$ for two solutions at same temperature and pressure is
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van't Haff - avogadro's law
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van't Hoff Boyle's law
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van't Hoff's law
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Henry's law
On the recently discovered $$10^{th}$$ planet it has been found that the gases follow the relationship $$Pe^{V/2} =nCT$$ where C is constant and other notations are as usual ( V in lit., P in atm and T in Kelvin). A curve is plotted between P and V at 500 K & 2 moles of gas as shown in the figure.
If a closed container of volume 200 lit. of $$O_2$$ gas (ideal gas) at 1 atm & 200 K is taken to the planet. Find the pressure of oxygen gas on the planet at 821 K in the same container:
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$$\cfrac { 10 } { e ^ { 100 } }$$
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$$\cfrac { 20 } { e ^ { 50 } }$$
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4 atm
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2 atm
In a mixture of $$A$$ and $$B$$, having vapour pressure of pure $$A$$ and pure $$B$$ as $$400mm\, Hg$$ and $$600mm\, Hg$$ respectively, mole fraction of $$B$$ in liquid phases is $$0.5$$. Calculate total vapour pressure and mole fraction of $$A$$ and $$B$$ in vapour phases.
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$$500, 0.4, 0.6$$
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$$500, 0.5, 0.5$$
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$$450, 0.4, 0.6$$
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$$450, 0.5, 0.5$$
Explanation
$$P_T = X_AP_{A^o} + X_BP_{B^o}$$
$$= 0.5\times 400+0.5\times 600$$
$$=500nm$$ of $$Hg$$
$$\dfrac{1}{P_T} = \dfrac{y_A}{P^o_A} + \dfrac{1-y_B}{P^o_B}$$
$$y_A = 0.6$$ & $$y_B = 1 -0.6 = 0.4$$
An ideal gas obeying kinetic theory of gases can be liquified, if?
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Its temperature is more than critical temperature $$T_c$$
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Its pressure is more than critical pressure $$P_c$$
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Its pressure is more than $$P_c$$ at a temperature less than $$T_c$$
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It cannot be liquified at any value of P and T
One monoatomic gas is expanded adibatically from $$2L$$ to $$10L$$ at $$1\ atm$$ external pressure find $$\triangle U$$ (in atm $$L$$)?
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$$-8$$
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$$0$$
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$$-66.7$$
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$$58.2$$
Explanation
Process is adiabatic $$\therefore Q = 0$$
$$\therefore \triangle U = W = -P_{ext} \triangle V$$
$$= -1 (10 - 2)\ atm \ L$$
$$= -8\ atm\ L$$.
The density of air 380K and 722 mm of Hg is 1g/$${ cm }^{ 3 }$$. If air is cooled to 100K and 1 atm the final density is:
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4
0%
5
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6
0%
7
When the air temperature is below freezing, the saturation vapor pressure over water is _______.
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Equal to zero
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Less than the saturation vapor pressure over ice
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Greater than the saturation vapor pressure over ice
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Equal to the saturation vapor pressure over ice
What is the total pressure (atm) in the chamber ?
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83.14
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831.4
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8.21
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None
Both stopcocks are opened and the system is again allowed to come at equilibrium. The pressure throughout the system is $$33.5$$ mm Hg. What do bulbs A, B and C contain?
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$$A:N_2(g), B:N_2(g), H_2O(s), C:N_2(g), H_2O(s), CO_2(s)$$
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$$A:N_2(g), B:N_2(g), H_2O(s), C:N_2(g), CO_2(s)$$
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$$A:N_2(g), B:N_2(g), H_2O(s), CO_2(s), C:N_2(g), H_2O(s), CO_2(s)$$
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$$A:N_2(g), B:N_2(g), H_2O(s), CO_2(g), C:N_2(g)m H_2O(s), CO_2(s)$$
Which of the following behaviour is true about the ideal binary liquid solution of liquids A and B if $$ P^0_A < P^0_B $$?
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plot of $$ P_{total}| vs X_A $$ in non linear
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plot of $$ P_{total}| vs X_B $$ is linear with +ve slope
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plot of $$ P_{total}| vs X_B $$ is linear with slope = 0
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plot of $$ P_{total}| vs X_B $$ is linear with -ve slope
The stopcock between A and B is opened and the system is allowed to come to equilibrium. The pressure in A and B is now $$0.21$$ atm.
How many moles of $$H_2O$$ are in the system?
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$$0.0075$$
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$$0.015$$
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$$0.05$$
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$$0.035$$
Explanation
No. of moles in flask A
Given:
P= 570 mm of Hg = 570/ 760 atm, V=1.642 L, T = 300 K
$$n=\dfrac{PV}{RT}=\dfrac{570\times 1.642}{760\times 0.082\times 300}$$
$$n=0.05$$ moles
$$n_{N_2}+n_{CO_2}+n_{H_2O}=0.05$$ moles
Upon opening the stop cock, the gases will dispose
In flask A: $$CO_2+N_2=(n_{CO_2}+n_{N_2})$$ ……$$(1)$$
In flask B: $$CO_2+N_2+H_2O=(n_{CO_2}+n_{H_2}+n_{H_2O})$$ ……….$$(2)$$
$$(2) - (1)$$
$$\Rightarrow n_{CO_2}+n_{N_2}+n_{H_2O}-(n_{CO_2}+n_{N_2})=n_{H_2O}$$
in flask A
no. of moles$$=\dfrac{0.21\times 1.642}{0.082\times 300}=0.014$$
in flask B
no. of moles$$=\dfrac{0.21\times 1.642}{0.082\times 200}=0.021$$
Total moles$$=(0.014+0.021)=n_{H_2O}$$
$$n_{H_2O}=0.05-0.035$$
$$=0.015$$ Option B.
In the case of positive deviation from an ideal gas?
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Interactions in molecules, $$\dfrac{PV}{nRT} > 1$$
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Interactions in molecules, $$\dfrac{PV}{nRT} < 1$$
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Finite size of molecules, $$\dfrac{PV}{nRT} > 1$$
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Finite size of molecule, $$\dfrac{PV}{nRT} < 1$$
Explanation
Condition for positive deviation of ideal gas
$$Z=\dfrac{PV}{nRT}$$
$$Z > 1$$ for positive deviation
Z(compressibility factor is also the ratio of actual molar mass of volume of gas to theoretical molar volume)
According to kinetic theory of gases:-
In positive deviation the interactions between molecules are very less, almost negligible.
The size of the molecules is negligible, but under the condition of deviation, from ideal gas equation, the size of molecules is not negligible.
$$\therefore$$ Size of molecules is finite, $$Z=\dfrac{PV}{nRT} > 1$$
Option C.
A vessel of volume $$10\ l $$ is evacuated by means of a piston air pump. One piston stroke captures the volume $$1\ l$$. The process is assumed to be isothermal and the gas ideal. The initial pressure of gas in the vessel was $$24.2$$ atm. Select the correct statement(s).
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The pressure of gas remained in the vessel after first stroke is $$22$$ atm
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The pressure of gas remained in the vessel after second stroke is $$20$$ atm
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The number of strokes needed to reduce the pressure in the vessel $$\eta$$ times in $$\dfrac{1 n\eta}{1n 1.1}$$
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The pressure of gas remained in the vessel after 'n' strokes is $$24.2\times \left(\dfrac{10}{11}\right)^n$$ atm
Explanation
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Benzene and toulene from an ideal solution, the vapour pressures of the benzene and toluene are 75 mm and 25 mm respectively at $$ 20^0 C $$ if the mole fractions of benzene and toluene in vapour are 0.75 and 0.25 respectively the vapour pressure of the ideal solution is
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62.5 mm
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50 mm
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30 mm
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40 m
The ratio between lowering of vapour pressure of solution and mole fraction of solute is equal to
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relative lowering of vapour pressure
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vapour pressure of pure solvent
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vapour pressure of solution
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molar mass of solvent
How many moles of $$CO_2$$ are in the system?
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$$0.0125$$
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$$0.015$$
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$$0.0225$$
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$$0.0375$$
Heptane and octane from ideal solution at 373 K the vapour pressure of the pure liquids are 106 kPa and 46 kpa respectively . what will be the vapour pressure, in bar of mixture of 30.0 g of heptane and 34.2 g of octane ?
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76 bar
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152 bar
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1.52 bar
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0.76 bar
For an ideal solution of A and B , $$ Y_A $$ is the mole fraction of A in the vapour phase at equilibrium . which of the following plot should be linear ?
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$$ P_{total} vs Y_A $$
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$$ P_{total} vs Y_B $$
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$$ \frac {1}{P_{total } } vs Y_A $$
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$$ \frac {1}{P_{total} } vs \frac {1}{Y_A } $$
If the pressure over the mixture at 300 K is reduced at what pressure does the first vapour form?
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40 mm Hg
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70 mm Hg
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100 Mm Hg
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199 mmHg
The vapour pressure of pure liquids A, B and C are 75, 22 and 10 torr , respectively. Which of the following is /are possible values of vapour pressure of binary or ternary solutions having equimolar amounts of these liquids? Assume ideal behaviour for all possible solutions
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53.5 torr
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35.67 torr
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48.5 torr
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16 torr
Explanation
$$P_{AB}= \cfrac{75+22}{2}=48.5$$ torr
$$P_{BC}= \cfrac{22+10}{2}=16$$ torr
$$P_{AC}= \cfrac{75+10}{2}=42.5$$ torr
$$P_{ABC}= \cfrac{75+22+10}{2}=35.67$$ torr
Option A, B and C are correct.
A mixture contains 1 mole of volatile liquid $$ A (P^0_A = 100 mm Hg ) $$ and 3 moles of volatile liquid$$ B (P^0_B = 80 mm Hg ) $$ if the solution behaves ideally, the total vapour pressure of the distillate is
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85 mm Hg
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85.88 mm Hg
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90 mm Hg
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92 mm Hg
Explanation
Mole fraction of A is $$X_A=\frac{1}{1+3}=0.25$$
Mole fraction of B is $$X_B=1-0.25=0.75$$
The expression for the total pressure of the solution is
$$P=P_A^0X_A+P_B^0X_B$$
Substitute values in the above expression
$$P=100\times 0.25+80\times 0.75=85$$ mm of Hg
Hence the correct answer is [A]
A liquid solution is formed by mixing 10 moles of anline and 20 moles of phenol at a temperature where the vapour phenol are 90 and 897 mm Hg, respectively . The possible vapour pressure of solution at that temperature is ?
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82 mm Hg
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88 mm Hg
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90 mm Hg
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94 mm Hg
An aqueous solution of sucrose is 0.5 molal. What is the vapour pressure of water above this solution ? The vapour pressure of pure water is 25.0 mm Hg at this temperature.
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24.8 mm Hg
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0.45 mm Hg
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2.22 mm Hg
0%
20.3 mm Hg
Explanation
Given that
$$ m = 0.5\;\;molal $$
V.P. (pure water) 25.0mm Hg
According to the
Raoult's law for dilute solution
$$\dfrac{P_0 - P_s}{P_0} = \dfrac{w}{m+W}\times 1000 = X_{sucrose}$$
here the 0.5 molal of solution means 0.5 mole in 1kg of water
Henc the mole of solvent = $$\dfrac{1000}{18}=55.5$$ [ the molecular mass of water molecules is 18]
then $$ \dfrac{25-P_s}{25} = \dfrac{0.5}{55.57+0.5}$$
$$ \Rightarrow\;1401.75 - 56.07P_s = 12.5 $$
$$\Rightarrow \; 56.07P_s = 1407.75 - 12.5$$
$$\Rightarrow \; P_S = \dfrac{1389.25}{\;\;56.07}$$
$$24.8\;mm \;Hg$$
The vapour pressure of the solution this composition is:
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$$ \sqrt { P^0_A .P^0_B } $$
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$$ ( P^0_A - P^0_B ) $$
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$$ ( P^0_A + P^0_B ) $$
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$$ 0.5 ( P^0_A + P^0_B ) $$
A liquid mixture of A and B (assume ideal solution ) is placed in a cylinder and piston arrangement . The piston is slowly pulled out isothermally so that the volume of liquid decreases and that of vapour increases. at the instant when the quantity of the liquid still remaining is negligibly small, the mole fraction of 'A' in the vapour is 0.4 .If $$ P^0_A = 0.4 atm , P^0_B = 1.2 atm $$ at this temperature, the total pressure at which the liquid has almost evaporated is ?
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0.667 atm
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1.5 atm
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0.8 atm
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0.545 atm
Liquids A and B from an ideal solution. A certain solution of A and B contains 25 mole percent of A whereas the vapours in the equilibrium with the solution at 298 K contains 50 mole per cent of A. The ratio of vapour pressures of pure A to that of pure B at 298 K , is
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1 :1
0%
3 : 1
0%
1 :3
0%
2 :1
Which of the following is the only incorrect informarion regarding the composition of the system at 0.51 bar pressure ?
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$$ X_A = 0.45 $$
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$$ Y_A = \frac {6}{17} $$
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$$ n_{A(liquid) } = \frac {12}{11} $$
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$$ n_{A(vapour) } = \frac {12}{11} $$
If the vapour are compressed slowly and isothermally , at what pressure , the first drop of liquid will appear ?
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0.4 bar
0%
0. 5 bar
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0.52 bar
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0.6 bar
One having high vapour pressure at temperature below its melting point is?
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Benzoic acid
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Salicylic acid
0%
Citric acid
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All of these
Explanation
Benzoic acid appears as a white crystalline solid.
Vapour pressures of benzoic acid were measured in the temperature range $$316\,K$$ to $$391 \,K$$ which is below its melting point $$395.1\, K$$. Hence correct answer is option A.
The vapour pressure of pure liquid A at $$ 80^0 C $$ is
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807.4 mm
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511.1 mm
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755.6 mm
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533.3 mm
The vapour pressure of pure liquid A at $$ 27^0 C $$ is
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60 torr
0%
48 torr
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49.26 torr
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61.58 torr
If the pressure is reduced further, at what presure does the trace of liquid disppear?
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57.14 mm Hg
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40 Mm Hg
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100 mm Hg
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66.67 mm Hg
At $$80^o C$$, the vapour pressure of pure liquid A is 520 mm Hg and that of pure liquid B is 1000 mm Hg. If a mixture solution of A and B boils at $$80^o C$$and 1 atm pressure, the amount of A in the mixture is :
(1 atm=760 mm Hg)
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34 mol%
0%
48 mol%
0%
50 mol%
0%
52mol%
The reading of pressure gauge at bubble point is :
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500
0%
600
0%
700
0%
None
Explanation
$$ X_A = 0.75 \quad X_B = 0.05 $$ above 500 mm Hg only liquid phase exists .
$$ P_{bubble} point = X_A P^0_A + X_B P^0_B $$
$$ P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B $$
$$ 0.75 +400 + 0.25 \times 800 = 500 mm $$
$$ y_A = 0.75 \quad \quad y_B = 0.5 $$
at dew point
$$ \frac {1}{P_T } = \frac {Y_A }{P^0A} + \frac {Y_B}{P^0_B } \Rightarrow \frac {1}{ P_T} = \frac { 0.75}{400} + \frac { 0.25}{800} = \frac { 1.5 + 0.25}{800} $$
$$ \Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg $$
Below dew point only vapor phase exists.
The reading of pressure gauge at which only vapour phase exists is
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0%
501
0%
457.14
0%
425
0%
525
Explanation
$$ X_A = 0.75 \quad X_B = 0.05 $$ above 500 mm Hg only liquid phase exists .
$$ P_{bubble} point = X_A P^0_A + X_B P^0_B $$
$$ P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B $$
$$ 0.75 +400 + 0.25 \times 800 = 500 mm $$
$$ y_A = 0.75 \quad \quad y_B = 0.5 $$
at dew point
$$ \frac {1}{P_T } = \frac {Y_A }{P^0A} + \frac {Y_B}{P^0_B } \Rightarrow \frac {1}{ P_T} = \frac { 0.75}{400} + \frac { 0.25}{800} = \frac { 1.5 + 0.25}{800} $$
$$ \Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg $$
Below dew point only vapor phase exists.
Correct gas equation is:
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$$\dfrac {V_{1} T_{2}}{P_{1} }= \dfrac {V_{2} T_{1}}{P_{2}}$$
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$$\dfrac {P_{1} V_{1}}{P_{2}V_{2} }= \dfrac {T_{1}} {T_{2}}$$
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$$\dfrac {P_{1} T_{2}}{V_{1} }= \dfrac {P_{2} V_{1}}{T_{2}}$$
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$$\dfrac {V_{1} V_{2}}{T_{1}T_{2} }= P_{1}P_{2}$$
Explanation
Ideal gas law equation:
$$PV = nRT$$
$$\dfrac {P_{1} V_{1}} {T_{1} }=\dfrac {P_{2}V_{2}}{T}$$
$$ \because$$ $$\dfrac{P_{1}V_{1}}{P_{2}V_{2}}=$$ $$\dfrac{T_{1}}{T_{2}}$$
Hence, option B is correct.
Vapour pressure of a solution is
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Directly proportional to the mole fraction of the solvent
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Inversely proportional to the mole fraction of the solute
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Inversely proportional to the mole fraction of the solvent
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Directly proportional to the mole fraction of the solute
Explanation
For solutions containing non - volatile solutes , the Raoult's law may be stated as at a given temperature, the vapour pressure of a solution containing non - volatile solute is directly proportional to the mole fraction of the solvent.
The vapour pressure of two micible liquid A and B are 300 and 500 mm of Hf respectively. in a flask , 2 moles of A are mixed with moles of B. further to the mixture , 32 g of an ionic non -volatile solute MCI (partially ionised, mol mass = 70 u) were also added. thus, the final vapour pressure of solution was found to be 420 mm and Hg. Then, identify the correct statement(s). (Assume the liquid mixture of A and B to behave ideally).
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The numerical value of relative lowering in vapour pressure upon addition of solute MCI is 1/15
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Th solute MCI is 25 % ionised in the above question.
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The solute MCI is 23.33 % ionised in the above question
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Upon addition of excess $$ Pb(NO_3)_2 $$ the number of moles of $$ PbCl_2 $$ precipitated is 2/ 35 .
Explanation
$$ P_0= X_A P^0_A + X_B P^0_B = \frac {2}{8} \times 300 + \frac {6}{7} \times 500 = 450$$ mm of Hg
Now RLVP = $$ \frac {P_0 -P_s}{P_0} = \frac { 450 - 420}{450} = \frac {1}{15} $$
Also RLVP = $$ ( \frac {in}{In +N } ) $$
$$ \frac {1}{15} = \frac { \frac { 32i}{70} }{ \frac {32i}{70} + 8 } $$
$$ \therefore i - 1.25 = ( 1+ ( 2-1 ) \alpha ). $$.
So $$ \alpha $$ = 0.25 ( or 25 %)
$$ \therefore n_{CI^- } $$ produced = $$ \frac {25}{100} \times \frac {32}{70} = \frac {4}{35} $$
$$ \therefore n_{PbCl_2 } $$ precipitated = $$ \frac {1}{2} \times \frac {4}{35} = \frac {2}{35} $$
Which of the following statement(s) is / are false?
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$$\Delta _r S$$ for $$\dfrac{1} {2} Cl_2(g) \rightarrow Cl(g)$$ is positive.
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$$\Delta E < 0$$ for combustion of $$CH_4(g)$$ in a sealed container with rigid adiabatic system
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$$\Delta G $$ is always zero for a reversible process in a closed system
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$$\Delta G ^ 0$$ for an ideal gas reaction is a function of pressure
Explanation
A) $$Cl_{2(g)} \rightarrow 2Cl_{(g)} $$ randomness is increasing, so entropy change is positive.
B) In closed container $$\Delta V=0$$ hence work done is zero. There is no heat exchange.Hence $$\Delta E=q+W=0$$
False statement
C) $$\Delta G$$ will be zero only when equilibrium is reached.
$$\therefore$$ false statement.
D) $$\Delta G^0=-RT\,ln\,k_{eq}$$
$$\therefore$$ not a function of pressure.
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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