Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

At 298 K , the K_(sp) of M_(2) SO_(4) is 3.2 xx 10^(-5) . The maximum concentration of SO_(4)^(2-) ion that could be attained in a saturated solution of this solid at 298 K is

Answer»

`3 XX 10^(-3)` M
`7 xx 10^(-2) M`
`2.89 xx 10^(-9) M`
`2 xx 10^(-2) M`

ANSWER :D
2.

At 298 K the K_c of reaction is 3.0xx10^14. At definate temp in reaction mixture [Cu^(2+)]=1.8xx10^(-2) M, [Ag^+]=3.0xx10^(-9) M is thisreaction in equilibrium ? In which direction reaction moves ?

Answer»

SOLUTION :`Q_c=2xx10^15 GT K_c=3.0xx10^14` So, REACTION is inreverse DIRECTION
3.

At 298 K, the [H_(2),O^(+)] of a solution is 2 xx 10^(-9)M. The nature of the solution is

Answer»

Acidic
Basic
Neutral
Can not be predicted

Solution :` [H^(+) ]= 2XX 10 ^(-9) lt 10 ^(-7), THEREFORE ` basic
4.

At 298 K temperature, there are acids and its K_a value are as under. HF(3.5xx10^(-4)),CH_3COOH (1.74xx10^(-5)),HClO(3.0xx10^(-8)),C_6H_5OH (1.3xx10^(-10)),HCOOH(1.8xx10^(-4))Arrange these acids in decreasing order of their acidic strength.

Answer»

Solution :If the VALUE of `K_a` is high then ACIDIC strength is high
HF `gt` HCOOH `gt CH_3COOH gt HClO gt C_6H_5OH`
`larr K_a` and acidic strength INCREASE `larr`
5.

At 298 K temperature, the sugar is added in open cup and stirred the solution then some amount of sugar remain undissolved then which type of equilibrium is it ?

Answer»

PHYSICAL equilibrium
Chemical equilibrium
Homogeneous cquilibrium
Hetrogeneous equilibrium

Answer :A::D
6.

At 298 K temperature the vapour pressure of water, acetone and ether are 234, 24.80 and 56.8 kPa respectively. Which are less vaporizable ?

Answer»

SOLUTION :ACETONE has LESS VAPOUR PRESSURE so it is less vapourisable.
7.

At 298 K temperature for this reaction 2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)) , K_p =4.0 xx10^24then at 500 K. What will be the value of K_p from these two value 2.5 xx 10^10 and 2.5 xx 10^(-10) ?

Answer»

SOLUTION :`K_p=2.5xx10^10`
8.

At 298 K temperature 4g H_(2) is filled in 500 mL vessel. Due to small hole in vessel, after some time, the pressure in vessel become 50 bar. Find the H_(2) of molecules which have escaped from the vessel ?(R=8.314xx10^(-2)" L bar mol"^(-1)K^(-1))(Note : Calculate mole, molecules using ofpV = nRT formula).

Answer»


ANSWER :`5.968xx10^(23)` MOLECULES
9.

At 298 K temp., the solubility product for AgCl is 1.5xx10^(-10) .Calculate its solubility in gram/Litre in pure water.

Answer»

SOLUTION :`1.75xx10^(-3) G L^(-1)`
10.

At 298 K temp. the K_(sp)of Mg(OH)_2 is 1.8 xx 10^(-11) . If the 0.1 M NaOH solution is added in it then what is the concentration of Mg (OH)_2 ? Calculate its solubility in water.

Answer»

Solution :solubility in WATER = `1.65xx10^(-4)` , solubility in NAOH = `1.8xx10^(-9)` M. THUS, adding NaOH solubility decreases
11.

At 298 K temp. the K_(sp) of CaF_2 is 1.7 xx 10^(-10). One person daily drinks 2.5 L saturated water by CaF_2. Then how much gram CaF_2 present in his body ? (Molecular mass of CaF_2 is "78 g mol"^(-1))

Answer»

SOLUTION :`S=3.489xx10^(-4)` M and 0.0683 G
12.

At 298 K, K_(p) for reaction N_(2)O_(4) (g) hArr 2NO_(2) (g) is 0.98. Predict whether the reaction is spontaneous or not.

Answer»

Solution :For the reaction, `N_(2)O_(4) (g) hArr 2NO_(2) (g), L_(P) = 0.98`
As we know that `Delta_(r) G^(Θ) = - 2.303 RT LOG K_(P)`
Here, `K_(P) = 0.98` i.e., `K_(P) lt 1` therefore, `Delta_(r) G^(@)` is positive, hence the reaction is non-spontaneous.
13.

At 298 K, K_(p) for the reaction NO_(2) O_(4(g)) hArr 2NO_(2(g))is 0.98. Predict whether the reaction is spontaneous or not.

Answer»

<P>

Solution :For the reaction, `N_(2) O_(4(g)) HARR 2NO_(2(g)) , K_(p) = 0.98`
As we know `Delta_(r) G^( Theta ) = -2.303 "RT logK"_(p)`
Here, `K_(p), = 0.98`
i.e., `K lt 1` therefore, `Delta_(r) G^( Theta )` is positive, hence the reaction is non-spontaneous.
14.

At 298 K for the reaction H_(2(g)) + J_(2(g)) hArr 2HJ_((g)) the K_c is 50.0. And for this reaction at any one state Q_c is 8.4 then reaction moves in which direction.

Answer»

SOLUTION :`Q_c 8.4 LT K_c 50.0` so REACTION take place in FORWARD direction and new EQUILIBRIUM is established.
15.

At 27^@C the reaction, C_6H_(6(l)) + 15/2 O_(2(g)) to 6CO_(2(g)) + 3H_2O((l)) proceeds spontaneously because the magnitude of

Answer»

`Delta H = T Delta S`
`Delta H GT T Delta S`
`Delta H LT T Delta S`
`Delta H gt 0, T Delta S lt 0`

ANSWER :B
16.

At 27^@C, the ratio of RMS velocities of ozone to oxygen is

Answer»

`SQRT(3//5)`
`sqrt(4//3)`
`sqrt(2//3)`
0.25.

Solution :`V_(RMS)=sqrt((3RT)/M)`
17.

At 27°C, the average speed of N_(2)molecules is xms4. At 927°, the speed will be_______ x ms^(-1)

Answer»

SOLUTION :`(C_1)/(C_2) = SQRT((T_1)/(T_2)) IMPLIES X/(C_2) = sqrt((300)/(1200)) = 1/2`.
18.

At 27^(@)C temperature in a 2L closed vessel 10 g H_(2) and 22g CO_(2) gases are filled. Find the partial pressure and total pressure of a mixture. (formula pV = nRT)

Answer»


Answer :`p_(H_(2))=62.35` bar, `p_(CO_(2))=6.24` bar, `p_("TOTAL")=68.59` bar
19.

At 27^(@)C temperature and 4 bar pressure CO_(2) is filled in 2 L vessel. Find the pressure if it is filled in 4L vessel at 77^(@)C temperature.

Answer»


ANSWER :2.33 BAR
20.

At 27^(@)C temperature 4 mole Cl_(2), 4 mole N_(2) and 2 mole O_(2) are filled in a 5 litre closed vessel. Find the total presure of gaseous mixture. (R=8.314xx10^(-2)"bar L mol"^(-1)K^(-1))

Answer»


ANSWER :49.88 BAR
21.

At 27^(@)C temperature and 1 bar pressure volume of gas is 25 L. If temperature becomes 77^(@)C at constant pressure then calculate volume of gas.

Answer»


ANSWER :29.17 L
22.

At 27^@C, one mole of an ideal gas ocmpressed isothermally and reversibly from a pressure of 2 atm to 1.0 atm, Choose the correct option from the following

Answer»

CHANGE in internal ENERGY is POSITIVE
Heat is negative
WORK done is `-965,84 `cal
All are incorrect

Answer :B
23.

At 27^(@)C latent heat of fusion of a compound is 2.7 xx 10^(3)J mol^(-1). Calculate the entropy change during fusion

Answer»


SOLUTION :`Delta S = (Delta H)/(MP) = (2.7 XX 10^(3))/(300) = 9`
24.

At 27^(@)C, average translational kinetic energies of the molecules in 8g of CH_(4),8g of O_(2) and 8g of He are overline(in_(1)),overline(in_(2)) and overline(in_(3)) respectively and total kinetic energies of the molecules in these gases are E_(1),E_(2) and E_(3) respectively. which of the following is true-

Answer»

`overline(in_(1))=overline(in_(2))=overline(in_(3))`
`overline(in_(3)) gt overline(in_(2)) gt overline(in_(1))`
`E_(1)=E_(2)=E_(3)`
`E_(2) LT E_(1) lt E_(3)`

ANSWER :A
25.

At 27^(@)C and one atmospheric pressure, a gas has volume V. What will be its volume at 177^(@)C and a pressure of 1.5 atmosphere ?

Answer»

SOLUTION :`{:("Initial conditions","FINAL conditions"),(V_(1)=V,V_(2)=?),(P_(1)=atm,P_(2)=1.5" atm"),(T_(1)=273+27=300 K,T_(2)=273+177=450 K):}`
APPLYING gas EQUATION, we have
`(1xxV)/(300)=(1.5xxV_(2))/(450):. V_(2)=(1xxVxx450)/(300xx1.5)=V:. "Volume of the gas"=V`
26.

At 27^@C, a cylinder of 20 L capacity contains three gases He, O_2 " and " N_2. Their masses are 0.502 g, 0.250 g and 1.00 g respectively. If all these gases behave ideally, calculate the partial pressure of each gas as well as the total pressure.

Answer»

Solution :Calculation of moles :
Moles of helium = `(0.502)/4 =0.125`
Moles of oxygen = `(0.250)/32 = 7.81xx10^(-3)`
Moles of NITROGEN = `(1.00)/28 = 3.57xx10^(-2)`
Calculation of partial PRESSURES :
`:. "" PV=nRT`
`:. ""P= (nRT)/V`
Given V=20L and T=27+273=300K
HENCE,
`P_(H_e)=(0.125xx0.0821xx300)/20 = 0.154` atm
`P_(O_2)= (7.81xx10^(-3)xx0.0821xx300)/20 = 9.62xx10^(-3)` atm
`P_(N_2) = (3.57xx10^(-2)xx0.0821xx300)/20 = 0.0440` atm
Calculation of total pressure :
ACCORDING to Dalton.s law of partial pressure,
`P_(mixture) = P_(He)+P_(O_2)+P_(N_2)=0.154+9.62xx10^(-3)+0.0440 = 0.208` atm
27.

At 273K, 2 moles of a gas is present in 22.4 lit vessel. The pressure of the gas is

Answer»

1 ATM 
2 atm 
4 atm 
3 atm 

Answer :B
28.

At 273 K temperature and 1 bar pressure volume of gas increases 20% then what temperature ?

Answer»


ANSWER :327 K
29.

A1 273 K and I atm, I Lof N_(2)O_(4(g)) decomposes to NO_(2(g)) as given, N_(2)O_(4(g)) 2NO_(2(g)), At equilibrium . original volume is 25% lessthan the exisiting volume percentage decomposition of N_(2)O_(4(g)) is thus,

Answer»

0.25
0.5
0.666
0.3333

Solution :PV=nRT, `V=(n)/(P)RT` at constant P & T , `V ALPHA n`
Thus, moles can be EXPRESSED in terms of volume
`underset(1-x)underset(1)(N_(2)O_(4(g))) HARR underset(2x)underset(0)(2NO_(2(g)))`
Total volume at equilibrium = (1+x)
Given `(75)/(100)(1+x)=1 IMPLIES x=(1)/(3)=0.333` = 33.3% dissociation
30.

At 273°C and 380 torr pressure, the density of a gas is 1.25kg//^(3) . So its density at STP in g// is

Answer»

5g/l
2.5 g/l
10 g/l
0.5 g/l

Answer :B
31.

At 27°C the ratio of the R.M.S. Velocities of ozone and oxygen molecules is

Answer»

` SQRT((3)/(5))`
` sqrt((1)/(4))`
`sqrt((4)/(3))`
`sqrt((2)/(3))`

ANSWER :D
32.

At 27^@C, one mole of an ideal gas exerted a pressure of 0.821 atmospheres. What is its volume in litres ? (R = 0.082 "lit-atm/mol"^(-1) K^(-1))

Answer»

300
30
0.3
3

Answer :B
33.

At 27^(@) C , a closed vessel contains a mixture of equal weights of helium (mol.wt = 4) , methane (mol.wt = 16) and sulphur dioxide (mol.wt = 64) . The pressure exerted by the mixture is 210 mm . If the partial pressures of helium , methane and sulphurdioxide are P_1 , P_2 and P_3 respectively , which one of the following is correct ?

Answer»

`P_3gt P_2 GT P_1`
`P_1gt P_2 gtP_3`
`P_1 gt P_3gt P_2`
`P_2gt P_3gt P_1`

ANSWER :B
34.

At 27^@ C, a closed vessel contains a mixture of equal weights of helium (mol. wt. = 4 ), methane (mol. wt. =16) and sulphur dioxide (mol. wt = 64). The pressure exerted by the mixture is 210 mm. If the partial pressures of helium, methane and sulphur dioxide are P_(1), P_(2) and P_(3) respectively, which one of the following is correct ?

Answer»

`P_(3) GT P_(2) P_(1)`
`P_(1) gt P_(2) gt P_(3)`
`P_(1) gt P_(3) gt P_(2)`
`P_(2) gt P_(3) gt P_(1)`

Answer :A
35.

At 25°C vapour pressure of water is 23 mm of Hg. If partial pressure of water vapour in air at 25°C is 18.4 mm of Hg, the percentage relative humidity in air is how many multiple of ten?

Answer»


SOLUTION :R.H % = `(18.4)/(23) XX 100 = 8`
36.

At 25^(@)C, the solubility product of Mg(OH)_(2) is 1.0xx10^(-11). At which pH will Mg^(2+) ions start precipitating in the form of Mg(OH)_(2) from a solution of 0.0001 M Mg^(2+) ions ?

Answer»

11
8
9
10

Solution :`K_(SP) `for `Mg(OH)_(2)=[Mg^(2+)][OH^(-)]^(2)`
`:. 1.0xx10^(-11)=(0.001)[OH^(-)]^(2)`
or `[OH^(-)]^(2)=10^(8) or [OH^(-)]=10^(-4)`
i.e, `pOH=4 or pH = 10`
37.

At 25^(@)C, the solubility product of Hg_(2)Cl_(2) is water is 3.2xx10^(-17) "mol"^(3) dm^(-9). What is the solubility of Hg_(2)Cl_(2) in water of 25^(@)C ?

Answer»

`1.2XX10^(-12) M`
`3.0xx10^(-6)M`
`2xx10^(-6)M`
`1.2xx10^(-16) M`

Solution :`{:(Hg_(2)Cl_(2),HARR,"Hg"_(2)^(2+),+,2Cl^(-)),(" "S,,S,,2S " "K_(sp)=S(2S)^(2)=4S^(3)):}`
`S=(K_(sp)//4)^(1//3)=(3.2xx10^(-17)//4)^(1//3)`
`=(8xx10^(-18))^(1//3)=2xx10^(-6)M`.
38.

At 25^(@)C, the ionisation constant of anilinium hydroxide is 4.6xx10^(-10). Taking ionic product of water as 1xx10^(-14), calculate (a) hydrolysis constant of anilinium chloride (b) the degree of hydrolysis and pH value of 0.2 molar solution of the salt.

Answer»


ANSWER :(a)`2.17xx10^(-5)(B) 1.042xx10^(-2)(C) 2.68`
39.

At 25^(@)C, the dissociation constant of a base, BOH is 1.0xx10^(-12). The concentration of hydroxy 1 ions in 0.01 M aqueous solution of the base would be :

Answer»

`1.0xx10^(-6) "mol " L^(-1)`
`1.0xx10^(-7) "mol " L^(-1)`
`2.0xx10^(-6) "mol " L^(-1)`
`1.0xx10^(-5)"mol " L^(-1)`

Solution :`BOHhArr B^(+) + OH^(-)`
`K=([B^(+)][OH^(-)])/([BOH]) = ([OH^(-)]^(2))/([BOH])`
or `[OH^(-)]=SQRT(K[BOH])=sqrt(10^(-12)(0.01))`
`=10^(-7)"mol" L^(-1)`
40.

At 25^(@)C, the combustion of 1 mol of liquid benzene, the heat of reaction at constant pressure is given by C_(6)H_(6)(l) + 7/2O_(2)(g) to 6 CO_(2)(g) + 3H_(2)O(l) Delta H = -780.980 cal Calculate the heat of reaction at constant volume.

Answer»

780.086kcal
`-782.470` kcal
`-390.043` kcal
390.043 kcal

Solution :`DELTAH = DELTAU + Deltan_(g) RT`
`Deltan_(g) = 6-7/2 = 2.5`
`DeltaU = DeltaH - Deltan_(g)RT`
` = -780980 -(2.5) xx 2 xx 298`
` = -782470` CAL
or = -782.470 kcal
41.

At 25^(@)C temperature and 760 mm Hg pressure volume of gas is 600 mL, Then calculate pressure when volume of this gas becomes 640 mL at 10^(@)C temperature. ((p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2)))

Answer»


ANSWER :676.6 MM HG
42.

At 25^(@)C, pH of a 0.01 M solution of a monobasic acid [HA] is 4. The correct statement(s) regarding HA and its given solution is (are)

Answer»

HA is a weak acid
The ionization constant `(K_(a))` of acid is approximately `10^(-6)` at `25^(@)C`.
Increasing the temperature of solution would CAUSE the `pH` to decrease.
Addition of `0.1 M HCl` solution would lower pH by increasing degree of ionization

Solution :N//A
43.

At 25^(@)C, K_(c ) for the reaction 3C_(2)H_(2(g))hArr C_(6)H_(6(g)) is 4.0. If the equilibrium concentration of C_(2)H_(2) is 0.5 mol. lit^(-1). What is the concentration of C_(6)H_(6)?

Answer»


Answer :`[C_(6)H_(6)]="0.5 MOL. Lit"^(-1)`
44.

At 25^(@)C, DeltaG^(@) for the processH_(2)O(l)iffH_(2)O(g) is 8.6 kJ. The vapour pressure of water at this temerapture, is nearly :

Answer»

24 torr
285 torr
32.17 torr
100 torr

Solution :At equilibrium `DELTAG^(@)=-RTlnK_(eq)`
where `K_("eq")=(P_(H_(2)O)//P^(@))" and "P^(@)=1" bar"`
45.

At 25^(@)C, Delta G^(0) for the process H_(2)O_((l)) hArr H_(2) O_((g)) is 8.6kJ. The vapour pressure of water at this temperature, is nearly

Answer»

24 torr
285 torr
32.17 torr
100 torr

Solution :`DELTA G^(0) = -RT ln K_(P) = - RT L n (P_(H_(2)O))`
`rArr P_(H_(2)O) = 0.03 ATM = 24` torr
46.

At 25^(@)C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10^(@)C and volume of the gas is 640 mL.

Answer»

<P>

Solution :Value of gases as follows
Initial temperature `(T_(1))=25^(@)C+273=298 K`
At High temperature `(T_(2))=10^(@)C+273=283 K`
Initial VOLUME `(V_(1))=600 ml`
High volume `(V_(2))=640 ml`
Initial Pressure `(p_(1))=760` mm of HG.
High Pressure `(p_(2))=(?)`
`(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2)) ""`.....(Combined GAS law)
`p_(2)=(("760 mm Hg")(600 mL))/((298 K))XX((283 K))/((640 mL))`
= 676.634 mm Hg `~~ 676.3` mm Hg.
47.

At 25^(@)C and 760 mm of Hg pressure of a gas occupies 600 ml volume. What will be its pressure at a height where temperature is 10^(@)C and volume of the gas is 640 mL ?

Answer»

<P>676.6 mm Hg
600 mm Hg
700 mm Hg
752 mm Hg

Solution :(a) According to combined gas law,`(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/T_(2)`
`rArr p_(2)=((760 mm Hg)XX(600 ML)xx(283 K))/((640 mL)xx(298K))=676.6 mm Hg`
48.

At 25^(@)C and 1 atm, a vessel contains 20 L of an ideal gas. If the volume of vessel increased to 40 L, the pressure exerted by gas in vessel will be "______________".

Answer»

2 ATM
1 atm
1.5 atm
0.5 atm

Answer :d
49.

At 25^(@)C and 760 mm of Hg pressure, a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10^(@)C and volume of the gas is 640 mL ?

Answer»

Solution :`{:("Initial conditions","Finalconditions"),(P_(1)=760 mm,P_(2)=?),(V_(1)=600 mL,V_(2)=640 mL),(T_(1)=25^(@)C=298 K,T_(2)=10^(@)C=283 K):}`
APPLYING gas equation (COMBINED gas LAW),
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))=((760 mm)(600 mL))/(298 K)=(P_(2)(640 mL))/(283 K)`
or`P_(2)=676.6" mm"`
50.

At 25^(@)C, a 0.01 mole sample of a gas is compressed from 4.0 L to 1.0 L at constant temperature. What is the work done for this process if the external pressure is 4.0 bar?

Answer»

`1.6 xx 10^(3) J`
`8.0 xx 10^(2) J`
`4.0 xx 10^(2) J`
`1.2 xx 10^(3) J`

Solution :`w=-4underset(1)OVERSET(4)intdV=+4xx3xx100, w=+1.2 xx10^(3)J`