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 :

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

$C$

For the adiabatic expansion of an ideal gas

0%
PV $^{\gamma}$ = constant
0%
TV $^{\gamma-1}$ = constant
0%
T $^{\gamma}$ P $^{1-\gamma}$ =constant
0%
All of these

A reaction follows the given concentration (M)- time graph. The rate for the reaction at 20 seconds will be

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

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

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

$P_1<P_2$

1070379_1199324_ans_e85b4422fa9142bbb25595b6f6ad2aec.image%2022

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?

0%
50
0%
70
0%
85
0%
80

One atmosphere is numerically equal to approximately?

0%
${ 10 }^{ 6 }{ dyne\quad cm }^{ -2 }$
0%
${ 10 }^{ 2 }{ dyne\quad cm }^{ -2 }$
0%
${ 10 }^{ 4 }{ dyne\quad cm }^{ -2 }$
0%
${ 10 }^{ 8 }{ dyne\quad cm }^{ -2 }$

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

$(p_{A}^{0}=300\ torr, p_{B}^{0}=800\ torr)$

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:

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)

1074580_1198189_ans_407c1a71e0ef4a478ebaafa0c45fee84.PNG

$200\ ml$ 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:

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

$O_{2}= 0.52\ atm$

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).

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 ?

0%
7.60 Torr
0%
76.00 Torr
0%
752.40 Torr
0%
759.00 Torr

Which of the following attraction is strongest?

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

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:

0%
$2 \times 10^{-4}\,atm$
0%
$5 \times 10^{-4}\,atm$
0%
$5 \times 10^{-2}\,atm$
0%
$5 \times 10^{-3}\,atm$

$n$ 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

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%
$\left[\dfrac{6V_{1}}{\pi N_{A}n}\right]^{1/3}$

An ideal gas is compressed in a closed container its U?

0%
Increases
0%
Decreases
0%
Remains same
0%
Both $(1)$ & $(2)$

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?

0%
Gay Lussac's law
0%
Law of definite proportion
0%
Law of multiple proportion
0%
Avogadro's law

Gay Lussac's law is not valid for which chemical reaction?

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%
$2SO_2(g)+O_2(g)\longrightarrow 2SO_3(g)$

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?

0%
48 mL
0%
60.0mL
0%
72 mL
0%
75.0 mL

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

$80\%$

0%
$2g$
0%
$4g$
0%
$8g$
0%
$10g$

Consider the followimg Vapour Pressure - Mole fraction graph. SP (Total vapour pressure of the resultant solution) is equal to

0%
PQ+RS
0%
PQ+QR+RS
0%
SR+SQ
0%
PQ+QR

Which of the following solutions will have the lowest vapour pressure?

0%
$0.1 M\ Glucose$
0%
$0.1 M\ NaCl$
0%
$0.1 M\ BaCl_2$
0%
$0.1 M Al_2 (SO_4)_3$

${40^ \circ }C$ 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

${40^ \circ }C$ ?

0%
$135$ and $254$ torr
0%
$119$ and $135$ tor
0%
$119$ and $254$ torr
0%
$334$ and $135$ torr

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

$\lim _{ { x }_{ A }\rightarrow 0 }{ \frac { { P }_{ B }^{ 0 } }{ { x }_{ B } } } \\$

0%
255
0%
115
0%
140
0%
135

A rigid vessel of volume 0.50

$m^3$ containing

$H_2$ at

$20.5^0 C$ and a pressure of

${ 611\times 10 }^{ 3 }$ Pa is connected to a second rigid vessel of volume 0.75

$m^3$ containing Ar at

$31.2^0 C$ at a pressure of

${ 611\times 10 }^{ 3 }$ Pa and temperature of

$14.5^0 C$ . What is the final pressure of the vessels?

0%
${ 2\times 10 }^{ 5 }$
0%
${ 3.22\times 10 }^{ 5 }$ Pa
0%
4840 Pa
0%
${ 4.84\times 10 }^{ 5 }$ Pa

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.

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?

0%
$3.444\ L$
0%
$4.344\ L$
0%
$4.798\ L$
0%
$1.212\ L$

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 -

0%
0.1
0%
0.2
0%
0.5
0%
1.0

How much pressure will be felt by a gas balloon which at

$100m$ depth in a sea from sea level.
[Pressure at ground level=

$10^5Pa,g=9.8m/s^2$ ]

0%
$100Pa$
0%
$98 Pa$
0%
$9.8\times 10^5Pa$
0%
$1.08\times 10^6Pa$

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 ?

0%
$54$
0%
$124$
0%
$180$
0%
$340$

$300ml$ of a gas at

$27^o$ C is cooled to

$7^oC$ at constant pressure, the final volume is

0%
$540ml$
0%
$135ml$
0%
$280ml$
0%
$350ml$

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:

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

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

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:

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,

$PV = n \times R \times T$

$200\times10=(0.5+x)\times R\times1000$

$2=(0.5+x)R$

$\dfrac{2}{R}=\dfrac{1}{2}+x$

$\dfrac{4}{R}-1=2x$

$\dfrac{4-R}{2R}=x$

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?

0%
0.346
0%
0.654
0%
0.5
0%
None of these

35 ml sample of hydrogen peroxide gives of 500ml of

$O_2$ at

$27^oC$ and 1 atm pressure. Volume strength of

$H_2O_2$ sample will be:-

0%
10 volume
0%
13 volume
0%
11 volume
0%
12 volume

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.)

0%
250 torr
0%
200 torr
0%
150 torr
0%
125 torr

${35}^{o}C$ , 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:

0%
A mixture of $100mL$ of acetone and $100mL$ of has a volume is $200mL$
0%
When acetone and are mixed at ${35}^{o}C$ , heat must be absorbed in order to produce a solution at ${35}^{o}C$
0%
When acetone and are mixed at ${35}^{o}C$ , heat is released
0%
There is negative deviation from Raoult's law

Vapour pressure increase with increase in :

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

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 :

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:

0%
$0.162$
0%
$0.675$
0%
$0.325$
0%
$0.486$

When a solution containing non-volatile solute is diluted with water,

0%
its vapour pressure increases
0%
its osmotic pressure increases
0%
its boiling point increases
0%
its freezing point increases

Gaseous mixture contains

$56 g$ of

$N_{2},44 g$ of

$CO_{2}$ and

$16g$ of

$CH_{4}$ . The total pressure of mixture is

$720mm$ of Hg. The partial pressure of methane is

0%
$75 atm$
0%
$160 atm$
0%
$180 atm$
0%
$215 atm$

The kinetic energy of

$1\ mole$ of gas is equal to-

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.

Henry's law constant K of

$CO_{2}$ in water at

$20^{\circ}C$ is

$3.0\times 10^{-2}molL^{-1}atm^{-1}$ . Calculate the mass of

$CO_{2}$ present in 100 L of soft drink bottled with a partial pressure of

$CO_{2}$ of 4 atm at the same temperature

0%
5.28 g
0%
12.0 g
0%
428 g
0%
528 g

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.

0%
$563$ and $469$
0%
$500$ adn $1250$
0%
$513$ and $494$
0%
$400$ and $600$

Pressure remaining constant, the volume of a given sample of gas at

${127}^{o}C$ will be doubled at

0%
${254}^{o}C$
0%
${527}^{o}C$
0%
$600K$
0%
${800}^{o}C$

Two flasks of equal volume, connected by a narrow tube of negligible volume, contain $2$ moles of $H_2$ gas at $1$ atm pressure and $300$ K. Now, one of the flasks are heated to $400$ K and the other is maintained at $300 K$ , then find final pressure:

0%
$\cfrac {8}{5}$ atm
0%
$\cfrac {5}{3}$ atm
0%
$\cfrac {7}{5}$ atm
0%
$\cfrac {8}{7}$ atm

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 13 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry States Of Matter Gases And Liquids Quiz 13 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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