MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Engineering Chemistry States Of Matter Quiz 13 - MCQExams.com
CBSE
Class 11 Engineering Chemistry
States Of Matter
Quiz 13
The solubility of a specific non-volatile salt is 4g in 100 g of water at $${25}^{0}$$C. If 2.0g, 4.0g and 6.0g of the salt added of 100g of water at $${25}^{0}$$C, in system X, Y and Z. The vapour pressure would be in the order:
Report Question
0%
$$Z>Y>X$$
0%
$$X>Y>Z$$
0%
$$Z>X=Y$$
0%
$$X>Y=Z$$
The pressure exerted by a mass of x mg resting on the area of 1.00 $${ cm }^{ 2 }$$ is 1.00 Pa, then x is
Report Question
0%
10.6
0%
10.3
0%
103
0%
10.2
According to kinetic theory of gases :
Report Question
0%
collisions are always elastic
0%
between collisions, the molecules move in straight lines with constant velocities
0%
only a small number of molecules have a very high velocity
0%
at of the above
$$200\ ml$$ of $$He$$ at $$0.66\ atm$$ pressure and $$400\ ml$$ of $${ O }_{ 2 }$$ at $$0.52\ atm$$ pressure are mixed in a $$400\ ml$$ vessel at $${ 25 }^{ \circ }C$$. The partial pressure of $$He$$ and $${ O }_{ 2 }$$ will be:
Report Question
0%
$$0.33\ and\ 0.56$$
0%
$$0.33\ and\ 0.52$$
0%
$$0.38\ and\ 0.56$$
0%
$$0.25\ and\ 0.45$$
Explanation
Partial pressure is the pressure of individual gas when it is placed in a required vessel or in final vessel.
Given :- Initially volume of $$He$$ is $$200\ ml$$ at $$0.66\ atm$$ after placing it in a vessel of volume $$400\ ml$$
$$P_{1} V_{1}= P_{2} V_{2}$$
$$.66 \times 200 = P_{2} \times 400$$
$$P_{2}= 0.33\ atm$$ ----------- (Partial pressure of $$He$$)
Initially $$O_{2}$$ is present at $$0.25\ atm$$ in $$ 400\ ml$$ volume and then it is placed in $$400\ ml$$ vessel hence the partial pressure of $$O_{2}$$ is $$0.52\ atm$$
Partial pressure of $$He=0.33\ atm$$
Partial pressure of $$O_{2}= 0.52\ atm$$
The reduction potential of hydrogen electrode will be positive if
Report Question
0%
$${P_{{H_2}}} = 2\;atm;\left[ {H^ + } \right] = 1\;M$$
0%
$${P_{{H_2}}} = 2.5\;atm;\left[ {H^ + } \right] = 1.5\;M$$
0%
$${P_{{H_2}}} = 2.5\;atm;\left[ {H^ + } \right] = 1\;M$$
0%
$${P_{{H_2}}} = 1\;atm;\left[ {H^ + } \right] = 2\;M$$
The vapour pressure of pure water at $$25^{\circ}C$$ is $$30\ mm$$. The vapour pressure of $$10\%\ (W/W)$$ glucose solution at $$25^{\circ}C$$ is :
Report Question
0%
$$31.5\ mm$$
0%
$$30.6\ mm$$
0%
$$29.67\ mm$$
0%
$$26.56\ mm$$
Explanation
$$10$$% solution $$\Rightarrow 10g$$ of glucose , $$90g$$ of water in $$100gm$$ of solution
So, number of moles of water in $$90gm$$=$$ \cfrac {90}{18}= 5$$ moles
Number of moles of glucose= $$\cfrac {10}{180}=0.056$$ moles
So, total moles= $$5+0.056=5.056$$ moles
Mole fraction of water= $$5/5.056=0.988$$
So, final vapour pressure= $$P\times X_{water}=30\times 0.980=29.6\ mm$$
Hence, the correct option is $$C$$
A reaction follows the given concentration (M)- time graph. The rate for the reaction at 20 seconds will be
Report Question
0%
410 MS
0%
810 MS
0%
210 MS
0%
710 MS
5.40 gm of an unknown gas at $${\text{2}}{{\text{7}}^{\text{o}}}{\text{C}}\,\,$$ occupies the same volume as 0.14 gm of hydrogen at $$1{{\text{7}}^{\text{o}}}{\text{C}}$$ and same pressure. The molecular weight of unknown gas is
Report Question
0%
79.8
0%
81
0%
79.2
0%
83
At point P and Q,the real gas deviation with respect to ideal gas is respectively
Report Question
0%
Positive,negative
0%
Positive, positive
0%
Negative,Positive
0%
Negative,Negative
Explanation
$$Z=\dfrac{PV}{RT}$$
at point $$Z=\dfrac{V_{seal}}{V_{seal}}=\dfrac{V_2}{V_1}$$
as from graph $$v_2>V_1$$
$$Z>1=+vE$$
at point $$Q$$
$$Z=\dfrac{P_{real}}{P_{idea}}=\dfrac{P_1}{P_2}$$
But $$P_1<P_2$$
The vapour pressure of pure benzene at a certain temperature is 200 mm Hg. At the same temperature the vapour pressure of a solution containing 2 g of non volatile non electrolyte solid in 78 g of benzene is 195 mm Hg. What is the molecular mass of solid?
Report Question
0%
50
0%
70
0%
85
0%
80
$$n$$ moles of Helium gas are placed in a vessel of volume $$V$$ Litre. At T K, if $$V_{1}$$ is free volume of Helium then diameter of He atom is
Report Question
0%
$$\left[\dfrac{3}{2}\dfrac{V_{1}}{\pi N_{A}n}\right]^{1/3}$$
0%
$$\left[\dfrac{3}{2}\dfrac{(V-V_{1})}{\pi N_{A}n}\right]^{1/3}$$
0%
$$\left[\dfrac{6(V-V_{1})}{\pi N_{A}n}\right]^{1/3}$$
0%
$$\left[\dfrac{6V_{1}}{\pi N_{A}n}\right]^{1/3}$$
Consider a binary mixture of volatile liquids. If at $$X_A=0.4$$ the vapour pressure of solution is $$580\ torr$$ then the mixture could be $$(p_{A}^{0}=300\ torr, p_{B}^{0}=800\ torr)$$
Report Question
0%
$$CHCl_3-CH_3COCH_3$$
0%
$$C_6H_5Cl-C_6H_5Br$$
0%
$$C_6H_6-C_6H_5CH_3$$
0%
$$nC_6H_{14}-nC_7H_{16}$$
The ratio of densities of a gas at two conditions of temperature and pressure is given by:
Report Question
0%
$$\dfrac{\rho_2}{\rho_1} = \left(\dfrac{p_2}{p_1} \right) \left(\dfrac{T_2}{T_1} \right)$$
0%
$$\dfrac{\rho_2}{\rho_1} = \left(\dfrac{p_2}{p_1} \right) \left(\dfrac{T_1}{T_2} \right)$$
0%
$$\dfrac{\rho_2}{\rho_1} = \left(\dfrac{p_1}{p_2} \right) \left(\dfrac{T_1}{T_2} \right)$$
0%
$$\dfrac{\rho_2}{\rho_1} = \left(\dfrac{p_1}{p_2} \right) \left(\dfrac{T_2}{T_1} \right)$$
Explanation
The density $$(S)$$ of an ideal gas is given as
$$S = \frac{ PM }{ RT }$$
Where,
$$P=$$ Pressure
$$T=$$ Temperature
$$R=$$ Universal gas content
$$M=$$ Molar Mass
Here, $$R$$ and $$M$$ are constant for a given gas
Therefore;
(refer image 1)
Calculate solubility (in moles / litre) of a saturated aqueous solution of $$Ag_3PO_4$$ if the vapour pressure of the solution becomes 750 torr at 373 K
(Assume molality=molarity).
Report Question
0%
$$2/15$$
0%
$$1/30$$
0%
$$10/54$$
0%
$$20/27$$
18 g of glucose $${ (C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 })$$ is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at $${ 100 }^{ \circ }C$$ is ?
Report Question
0%
7.60 Torr
0%
76.00 Torr
0%
752.40 Torr
0%
759.00 Torr
Gay Lussac's law is not valid for which chemical reaction?
Report Question
0%
$$2NO(g)+O_2(g)\longrightarrow 2NO_2(g)$$
0%
$$2HI(g)\longrightarrow H_2(g)+I_2(g)$$
0%
$$4B(s)+3O_2(g)\longrightarrow 2B_2O_3(s)$$
0%
$$2SO_2(g)+O_2(g)\longrightarrow 2SO_3(g)$$
Which of the following attraction is strongest?
Report Question
0%
0%
0%
0%
Liquids $$A$$ and $$B$$ form an ideal solution. At $$30^o$$ C, the total vapour pressure of a solution containing $$1$$ mol of $$A$$ and $$2$$ mols of $$B$$ is $$250$$ mm $$Hg$$. The total vapour pressure becomes $$300$$ mm $$Hg$$ when $$1$$ more mol of A is added to the first solution. The vapour pressures of pure $$A$$ and $$B$$ at the same temperature are
Report Question
0%
$$450, 150$$ mm $$Hg$$
0%
$$150, 450$$ mm $$Hg$$
0%
$$250, 300$$ mm $$Hg$$
0%
$$125, 150$$ mm $$Hg$$
The equilibrium constant for the following equilibrium is given at $$0^0C$$:
$$Na_2HPO_4.12H_2O(s)\rightarrow Na_2HPO_4.7H_2O(s)+5H_2O(g); K_p=31.25 \times 10^{-13}$$.
The vapor pressure of water is:
Report Question
0%
$$2 \times 10^{-4}\,atm$$
0%
$$5 \times 10^{-4}\,atm$$
0%
$$5 \times 10^{-2}\,atm$$
0%
$$5 \times 10^{-3}\,atm$$
An ideal gas is compressed in a closed container its U?
Report Question
0%
Increases
0%
Decreases
0%
Remains same
0%
Both $$(1)$$ & $$(2)$$
Explanation
When an ideal gas is compressed in a closed container, its internal energy U increases. Compressing an ideal gas increases its temperature and its internal energy increases since U = f(t) for an ideal gas.
45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below :
$$2N_{2(g)}+O_{2(g)}. \rightarrow . 2N_2O_(g)$$
Which law is being obeyed in this experiment?
Report Question
0%
Gay Lussac's law
0%
Law of definite proportion
0%
Law of multiple proportion
0%
Avogadro's law
Which of the following mixture of gases does obey Dalton's Law of partial pressure ?
Report Question
0%
$$Cl_2$$ and $$SO_2$$
0%
$$CO_2$$ and $$He$$
0%
$$O_2$$ and $$CO_2$$
0%
$$N_2$$ and $$O_2$$
A mixture of 50.ml of $$N{H_3}$$ and 60.0 ml of $${O_2}$$ gas react as
$$4N{H_3}(g) + 5{O_2}(g)\xrightarrow{{}}4NO + 6{H_2}O(g)$$
If all the gases are at the same temperature and the reaction continues until one of the gases is completely consumed, what volume of water vapour is produced?
Report Question
0%
48 mL
0%
60.0mL
0%
72 mL
0%
75.0 mL
Two liquids A and B have $${P_A}^0:{P_B}^0$$ = 1:3 at a certain temperature. If the mole fraction ratio $${x_A}:{x_B} = $$ 1:3, the mole fraction of A in vapour in equilibrium with the solution at the given temperature is -
Report Question
0%
0.1
0%
0.2
0%
0.5
0%
1.0
Calculate the mass of a non - volatile solute (molar mass $$40\,\,g mol^{-1}$$) which should be dissolved in $$114g$$ octane to reduce its vapour pressure to $$80\%$$
Report Question
0%
$$2g$$
0%
$$4g$$
0%
$$8g$$
0%
$$10g$$
Initial temperature of an ideal gas is $${\text{7}}{{\text{5}}^{\text{o}}}{\text{C}}$$ . At what temperature, the sample of neon gas would be heated to double its pressure, if the initial volume of gas is reduced by 15%?
Report Question
0%
$${\text{31}}{{\text{9}}^{\text{o}}}{\text{C}}$$
0%
$${\text{59}}{{\text{2}}^{\text{o}}}{\text{C}}$$
0%
$${\text{12}}{{\text{8}}^{\text{o}}}{\text{C}}$$
0%
$${\text{6}}{{\text{0}}^{\text{o}}}{\text{C}}$$
Explanation
$$T_{1} = 75^{o}C = 75 + 273 = 348 K$$
Let $$V_{1} = V $$
and $$ V_{2} = {85V}/{100}$$
$${V_{1}}/{T_{1}} = {V_{2}}/{T_{2}}$$
$$T_{2} = \dfrac{V_{2}T_{1}}{V_{1}} = \dfrac{85V \times 348}{100 \times V}$$
$$= 295.8 K$$
$$P_{1} = P $$
$$P_{2} = 2P$$
$$T_{1} = 295.8 K $$
$$T_{2} = ?$$
$${P_{1}}/{T_{1}} = {P_{2}}/{T_{2}}$$
$$T_{2} =\dfrac {P_{2}T_{1}}{P_{1}} = \dfrac{2P \times 295.8}{P}$$
$$= 591.6 K$$ $$= {591.6 - 273} .^{o}C = 318.6^{o}C$$ $$~~ 319^{o}$$
Which of the following solutions will have the lowest vapour pressure?
Report Question
0%
$$0.1 M\ Glucose$$
0%
$$0.1 M\ NaCl$$
0%
$$0.1 M\ BaCl_2$$
0%
$$0.1 M Al_2 (SO_4)_3$$
At $${40^ \circ }C$$ the vapour pressure ( in torr ) of a mixture of methyl alcohol and ethyl alcohol is represented by $$P = 199x + 135$$
(where $$x$$ is the mole fraction of methyl alcohol)
What are the vapour pressures of pure methyl alcohol and pure ethyl alcohol at $${40^ \circ }C$$ ?
Report Question
0%
$$135$$ and $$254$$ torr
0%
$$119$$ and $$135$$ tor
0%
$$119$$ and $$254$$ torr
0%
$$334$$ and $$135$$ torr
Explanation
Correct option is D
Vapour Pressure of methyl alcohol and ethyl alcohol solution is represented by $$P=115{ x }_{ A}+140$$. Where $${ x }_{ A }$$ is the mole fraction of methyl alcohol. The value of $$\lim _{ { x }_{ A }\rightarrow 0 }{ \frac { { P }_{ B }^{ 0 } }{ { x }_{ B } } } \\ $$
Report Question
0%
255
0%
115
0%
140
0%
135
Calculate the vapour pressure of aqueous 0.1 m glucose solution at 300 K temperature, the vapour pressure of water is 0.03 bar at 300 K temperature.
Report Question
0%
0.5 bar
0%
0.29 bar
0%
0.3 bar
0%
0.03 bar
$$5.0\ L$$ water placed in a closed room of volume $$2.5\times 10^{4}L$$ having temperature $$300\ K$$. If vapour pressure of water is $$27.0\ mm$$ and density is $$0.990\ g/cm^{3}$$ at this temperature, how much water is left in liquid state?
Report Question
0%
$$3.444\ L$$
0%
$$4.344\ L$$
0%
$$4.798\ L$$
0%
$$1.212\ L$$
At a certain temperature pure liquid A and liquid B have vapour pressures 10 torr and 37 torr respectively. For a certain ideal solution of A and B, the vapour in equilibrium with the liquid has the components A and B in the partial pressure ratio $$P_A : P_B=1 : 7 $$. What is the mole fraction of A in the solution?
Report Question
0%
0.346
0%
0.654
0%
0.5
0%
None of these
The energy of absorbed by each molecule $$({A_2})$$ of a substance is $$4.4 \times {10^{ - 19}}J$$ and bond energy per molecule is $$4.0\times {10^{ - 19}}J$$. The kinetic energy of the molecule per atom will be
Report Question
0%
$$2.2 \times {10^{ - 19}}J$$
0%
$$2.0 \times {10^{ - 19}}J$$
0%
$$4.0 \times {10^{ - 20}}J$$
0%
$$2.0 \times {10^{ - 20}}J$$
Vapour pressure increase with increase in :
Report Question
0%
Concentration of solution containing non-volatile soulute
0%
Temperature up to boiling point
0%
Temperature upto triple point
0%
Altitude of the concerned place of boiling
The vapour pressure of the solution of $$5\,g$$ on non-electrolyte in $$100\,g$$ of water at a particular temperature is $$2950\,N/{m^2}$$. The vapour pressure of water is $$3000\,N/{m^2}$$. The mocular mass of the solute is ?
Report Question
0%
$$54$$
0%
$$124$$
0%
$$180$$
0%
$$340$$
The kinetic energy of two moles of $$ N_2 $$ at $$ 27^oC$$ is $$(R=8.314 \ JK^{-1} mol^{-1}) $$
Report Question
0%
5491.6 J
0%
6491.6 J
0%
7482.6 J
0%
8882.4 J
The mass of a non-volatile non-electrolyte solute (molar mass$$=50\ g\ mol^ {-1}$$) needed to be dissolved in $$114\ g$$ octane to reduce its vapour pressure to $$75\%$$ is:
Report Question
0%
$$37.5\ g$$
0%
$$75\ g$$
0%
$$150\ g$$
0%
$$50\ g$$
At a certain temperature, the vapour pressure of water is $$50\ mm.$$ The relative lowering of vapour pressure of a solution containing $$36\ g$$ of glucose in $$900\ g$$ of water is
Report Question
0%
$$0.1$$
0%
$$0.5$$
0%
$$0.004$$
0%
$$3.75$$
The vapour pressure of $$1$$ molal glucose solution at $$100^{o}C$$ will be
Report Question
0%
$$760\ mmHg$$
0%
$$76.0\ mmHg$$
0%
$$1\ atm$$
0%
$$746.32\ mmHg$$
$$0.5$$ moles of gas $$A$$ and $$x$$ moles of gas $$B$$ exert a pressure of $$200$$ Pa in a a container of volume $$10 m^3$$ at $$1000$$ K. Given $$R$$ is the gas constant in $$JK^{-1}$$ mol$$^{_1}$$, $$x$$ is:
Report Question
0%
$$\dfrac{2R}{4+ 12}$$
0%
$$\dfrac{2R}{4- 12}$$
0%
$$\dfrac{4-R}{2R} $$
0%
$$\dfrac{4+R}{2R} $$
Explanation
Given that
$$n_T= (0.5 +x);\quad T=1000K; \quad V=10m^3; \quad P =200Pa$$
Now,
$$PV = n \times R \times T$$
$$200\times10=(0.5+x)\times R\times1000$$
$$2=(0.5+x)R$$
$$\dfrac{2}{R}=\dfrac{1}{2}+x$$
$$\dfrac{4}{R}-1=2x$$
$$\dfrac{4-R}{2R}=x$$
When a solution containing non-volatile solute is diluted with water,
Report Question
0%
its vapour pressure increases
0%
its osmotic pressure increases
0%
its boiling point increases
0%
its freezing point increases
An ideal solution formed by mixing two liquids 'A' and 'B' find the pressure at which $${X_A} = {Y_B}\ ({P_A}^0 = 400\ torr,{P_B}^0 = 100\,torr)$$
$$({X_A}$$ and $${Y_B}$$ are mole fraction of A and B in liquid and vapour forms respectively at equilibrium.)
Report Question
0%
250 torr
0%
200 torr
0%
150 torr
0%
125 torr
At $${35}^{o}C$$, the vapour pressure of $$CS_2$$ is $$512mm$$ $$Hg$$, and acetone is $$344mm$$ $$Hg$$. A solution of
$$CS_2$$
and acetone in, which the mol fraction of
$$CS_2$$
is $$0.25$$, has a total vapour pressure of $$600mm$$ $$Hg$$. Which of the following statements is/are correct:
Report Question
0%
A mixture of $$100mL$$ of acetone and $$100mL$$
$$CS_2$$
of has a volume is $$200mL$$
0%
When acetone and
$$CS_2$$
are mixed at $${35}^{o}C$$, heat must be absorbed in order to produce a solution at $${35}^{o}C$$
0%
When acetone and
$$CS_2$$
are mixed at $${35}^{o}C$$, heat is released
0%
There is negative deviation from Raoult's law
Vapour pressure of pure 'A' is 70 mm of Hg at $$25^0C$$ it forms an ideal solution is with 'B' in which mole fraction of A is 0.If the vapour pressure of the solution is 84 mm is Hg at $$25^0C$$, the vapour pressure of pure 'B' at $$25^0C$$ is :
Report Question
0%
56 mm Hg
0%
70 mm Hg
0%
140 mm Hg
0%
28 mm Hg
A solution is prepared by mixing $${\text{8}}{\text{.5}}\,{\text{g}}$$ of $${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_2}$$ and $$11.95\,{\text{g}}$$ of $${\text{CHC}}{{\text{l}}_{\text{3}}}$$. If vapour pressure of $${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$$ and $${\text{CHC}}{{\text{l}}_3}$$ at $${\text{298}}\,{\text{K}}$$ are $$415$$ and $$200\,{\text{mm}}$$ Hg respecetively, then mole fraction of $${\text{CHC}}{{\text{l}}_{\text{3}}}$$ in vapour form is:
Report Question
0%
$$0.162$$
0%
$$0.675$$
0%
$$0.325$$
0%
$$0.486$$
Two liquids A and B form ideal solution.At 300K,the vapour pressure of solution containing 1 mol of A and 3 mol of B is 550mm Hg.At the same temperature,if one more mole of B is added to this solution,the vapour pressure of the solution increases by 10mm Hg.Determine the vapour pressures of A and B in their pure states.
Report Question
0%
$$563$$ and $$469$$
0%
$$500$$ adn $$1250$$
0%
$$513$$ and $$494$$
0%
$$400$$ and $$600$$
The volume occupied by 4.4 gram of $$C{O_2}$$ at STP is
Report Question
0%
2.85 liter
0%
2.24 liter
0%
2.6 liter
0%
2.64 liter
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)
Report Question
0%
34 mol%
0%
48 mol%
0%
50 mol%
0%
52mol%
The kinetic energy of $$1\ mole$$ of gas is equal to-
Report Question
0%
$$\dfrac {3}{2}RT$$
0%
$$\dfrac {3}{2}KT$$
0%
$$\dfrac {RT}{2}$$
0%
$$\dfrac {2R}{3}$$
Explanation
Kinetic enedy of gaseous molecule is calculaed using following formula.
$$KE= \frac{3}{2}\times nRT$$.
Above formula is given for n moles
So for 1 mole $$n=1$$
$$KE= \frac{3}{2}\times 1\times RT$$.
$$KE= \frac{3}{2}\times RT$$.
Hence option A is correct.
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
Report Question
0%
hydrogen
0%
methane
0%
oxygen
0%
soldier oxide
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Engineering Chemistry Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
